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All questions of Work, Energy and Power for JEE Exam
An athlete in the olympic games covers a distance of 100 m in 10s. His kinetic energy can be estimated to be in the range- a)200 J - 500 J
- b) 2 × 105J - 3 × 105J
- c)20, 000 J - 50,000 J
- d)2,000 J - 5, 000 J
Correct answer is option 'D'. Can you explain this answer?
An athlete in the olympic games covers a distance of 100 m in 10s. His kinetic energy can be estimated to be in the range
a)
200 J - 500 J
b)
2 × 105J - 3 × 105J
c)
20, 000 J - 50,000 J
d)
2,000 J - 5, 000 J
|
Samridhi Choudhury answered |
The kinetic energy (KE) of an object is given by the equation KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.
In this case, we are given the distance covered (100 m) and the time taken (10 s). We can use these values to find the velocity (v) of the athlete.
Velocity (v) = Distance / Time = 100 m / 10 s = 10 m/s
Now, let's assume the mass of the athlete is 70 kg (a typical mass for an athlete).
Using the equation for kinetic energy, we can calculate the estimated range of the athlete's kinetic energy.
KE = 1/2 * m * v^2
KE = 1/2 * 70 kg * (10 m/s)^2
KE = 1/2 * 70 kg * 100 m^2/s^2
KE = 3500 J
Therefore, the estimated range for the athlete's kinetic energy is approximately 3500 J.
Since none of the given answer choices include this value, it seems that there may be a mistake in the question or the answer choices provided.
In this case, we are given the distance covered (100 m) and the time taken (10 s). We can use these values to find the velocity (v) of the athlete.
Velocity (v) = Distance / Time = 100 m / 10 s = 10 m/s
Now, let's assume the mass of the athlete is 70 kg (a typical mass for an athlete).
Using the equation for kinetic energy, we can calculate the estimated range of the athlete's kinetic energy.
KE = 1/2 * m * v^2
KE = 1/2 * 70 kg * (10 m/s)^2
KE = 1/2 * 70 kg * 100 m^2/s^2
KE = 3500 J
Therefore, the estimated range for the athlete's kinetic energy is approximately 3500 J.
Since none of the given answer choices include this value, it seems that there may be a mistake in the question or the answer choices provided.
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table ?- a)12 J
- b)3.6 J
- c)7.2 J
- d)1200 J
Correct answer is option 'B'. Can you explain this answer?
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table ?
a)
12 J
b)
3.6 J
c)
7.2 J
d)
1200 J
|
Upasana Ahuja answered |
Given:
Length of the chain (L) = 2 m
Length hanging freely (h) = 60 cm = 0.6 m
Total mass of the chain (m) = 4 kg
To find: Work done in pulling the entire chain on the table
Work done can be calculated using the formula:
Work = Force × Distance
1. Calculating the force:
The force acting on the chain is equal to the weight of the hanging part of the chain.
The weight of the hanging part can be calculated using the formula:
Weight = mass × acceleration due to gravity
The mass of the hanging part can be calculated using the formula:
Mass = (Length of the hanging part / Total length) × Total mass
Mass of the hanging part (m_h) = (0.6 / 2) × 4 = 1.2 kg
Weight of the hanging part (W) = m_h × g, where g is the acceleration due to gravity (approximated to 9.8 m/s²)
W = 1.2 × 9.8 = 11.76 N
2. Calculating the distance:
The distance over which the force is applied is equal to the length of the hanging part of the chain.
Distance (d) = Length of the hanging part = 0.6 m
3. Calculating the work done:
Work = Force × Distance
Work = W × d
Work = 11.76 × 0.6
Work = 7.056 J
Therefore, the work done in pulling the entire chain on the table is 7.056 J, which is approximately equal to 7.2 J.
Hence, option C is the correct answer.
Length of the chain (L) = 2 m
Length hanging freely (h) = 60 cm = 0.6 m
Total mass of the chain (m) = 4 kg
To find: Work done in pulling the entire chain on the table
Work done can be calculated using the formula:
Work = Force × Distance
1. Calculating the force:
The force acting on the chain is equal to the weight of the hanging part of the chain.
The weight of the hanging part can be calculated using the formula:
Weight = mass × acceleration due to gravity
The mass of the hanging part can be calculated using the formula:
Mass = (Length of the hanging part / Total length) × Total mass
Mass of the hanging part (m_h) = (0.6 / 2) × 4 = 1.2 kg
Weight of the hanging part (W) = m_h × g, where g is the acceleration due to gravity (approximated to 9.8 m/s²)
W = 1.2 × 9.8 = 11.76 N
2. Calculating the distance:
The distance over which the force is applied is equal to the length of the hanging part of the chain.
Distance (d) = Length of the hanging part = 0.6 m
3. Calculating the work done:
Work = Force × Distance
Work = W × d
Work = 11.76 × 0.6
Work = 7.056 J
Therefore, the work done in pulling the entire chain on the table is 7.056 J, which is approximately equal to 7.2 J.
Hence, option C is the correct answer.
A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to- a)v
- b)v2
- c)v3
- d)v4
Correct answer is option 'C'. Can you explain this answer?
A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to
a)
v
b)
v2
c)
v3
d)
v4
|
Amrutha Chaudhary answered |
A spring of spring constant 5 × 103 N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm is- a)12.50 N-m
- b)18.75 N-m
- c)25.00 N-m
- d)6.25 N-m
Correct answer is option 'B'. Can you explain this answer?
A spring of spring constant 5 × 103 N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm is
a)
12.50 N-m
b)
18.75 N-m
c)
25.00 N-m
d)
6.25 N-m
|
Amrita Sharma answered |
k = 5*103 N/m
A spring of force-constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of- a)(2/3)k
- b)(3/2)k
- c)3 k
- d)6 k
Correct answer is option 'B'. Can you explain this answer?
A spring of force-constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of
a)
(2/3)k
b)
(3/2)k
c)
3 k
d)
6 k
|
Stuti Chopra answered |
Given:
A spring of force-constant k is cut into two pieces such that one piece is double the length of the other.
To find:
The force constant of the longer piece of the spring.
Explanation:
Let the original spring be of length L and the shorter piece be of length x. Therefore, the longer piece will be of length 2x.
The force constant, also known as the spring constant, is defined as the force required to stretch or compress a spring by a certain amount. It is denoted by k and is a measure of the stiffness of the spring.
The force constant of a spring is given by the formula:
F = kx
Where F is the force applied, k is the force constant, and x is the displacement.
Applying the formula:
Let's consider the shorter piece of the spring:
F_shorter = k_shorter * x
Similarly, for the longer piece:
F_longer = k_longer * 2x
Since the force constant is directly proportional to the force, we can write:
F_longer = 2 * F_shorter
Substituting the values, we get:
k_longer * 2x = 2 * k_shorter * x
k_longer = 2 * k_shorter
Conclusion:
Therefore, the force constant of the longer piece of the spring is twice the force constant of the shorter piece.
So, the correct answer is option B) (3/2)k.
A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is- a)MgL
- b)MgL/3
- c)MgL/9
- d)MgL/18
Correct answer is option 'D'. Can you explain this answer?
A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is
a)
MgL
b)
MgL/3
c)
MgL/9
d)
MgL/18
|
Devika Banerjee answered |
The hanging part of the chain which is to be pulled up can be considered as a point mass situated at the centre of the hanging part. The equivalent diagram is drawn.
Note : The work done in bringing the mass up will be equal to the change in potential energy of the mass.
Note : The work done in bringing the mass up will be equal to the change in potential energy of the mass.
W = Change in potential energy
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle.
The motion of the particle takes place in a plane. It follows that :- a)its velocity is constant
- b) its acceleration is constant
- c)its kinetic energy is constant.
- d)it moves in a circular path.
Correct answer is option 'C,D'. Can you explain this answer?
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle.
The motion of the particle takes place in a plane. It follows that :
The motion of the particle takes place in a plane. It follows that :
a)
its velocity is constant
b)
its acceleration is constant
c)
its kinetic energy is constant.
d)
it moves in a circular path.
|
Dipika Choudhury answered |
When the force is perpendicular to the velocity and constant in magnitude then the force acts as a centripetal force, and the body moves in a circular path. The force is constant in magnitude, this show the speed is not changing and hence kinetic energy will remain constant.
Note : The velocity changes continuously due to change in the direction. The acceleration also changes continuously due to change in direction.
When a rubber-band is stretched by a distance x, it exerts restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is:- a)aL2 + bL3
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
When a rubber-band is stretched by a distance x, it exerts restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is:
a)
aL2 + bL3
b)
c)
d)
|
|
Akash Kumar answered |
Work done due to a variable force is the integration of F.dx with the limit of initial length to final length.so by keeping the value of force in terms of x and integrating the given equation.you will get your answer.
try it if you did not get the answer reply me.
try it if you did not get the answer reply me.
A particle of mass 100g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is- a)–0.5 J
- b)–1.25 J
- c)1.25 J
- d)0.5 J
Correct answer is option 'B'. Can you explain this answer?
A particle of mass 100g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is
a)
–0.5 J
b)
–1.25 J
c)
1.25 J
d)
0.5 J
|
|
Navya Pillai answered |
To calculate the work done by the force of gravity, we need to determine the change in potential energy of the particle as it goes up.
The potential energy of an object near the surface of the Earth is given by the equation:
Potential Energy = mass * acceleration due to gravity * height
In this case, the mass of the particle is 100g, which is equivalent to 0.1 kg. The acceleration due to gravity is approximately 9.8 m/s^2. The height the particle reaches is determined by its initial velocity and the time it takes to reach its maximum height.
Using the equation for the height of an object in free fall:
Height = (initial velocity)^2 / (2 * acceleration due to gravity)
Height = (5 m/s)^2 / (2 * 9.8 m/s^2) = 1.275 m
Now we can calculate the change in potential energy:
Change in Potential Energy = mass * acceleration due to gravity * change in height
Change in Potential Energy = 0.1 kg * 9.8 m/s^2 * 1.275 m = 1.24 J
Therefore, the work done by the force of gravity during the time the particle goes up is 1.24 Joules.
The potential energy of an object near the surface of the Earth is given by the equation:
Potential Energy = mass * acceleration due to gravity * height
In this case, the mass of the particle is 100g, which is equivalent to 0.1 kg. The acceleration due to gravity is approximately 9.8 m/s^2. The height the particle reaches is determined by its initial velocity and the time it takes to reach its maximum height.
Using the equation for the height of an object in free fall:
Height = (initial velocity)^2 / (2 * acceleration due to gravity)
Height = (5 m/s)^2 / (2 * 9.8 m/s^2) = 1.275 m
Now we can calculate the change in potential energy:
Change in Potential Energy = mass * acceleration due to gravity * change in height
Change in Potential Energy = 0.1 kg * 9.8 m/s^2 * 1.275 m = 1.24 J
Therefore, the work done by the force of gravity during the time the particle goes up is 1.24 Joules.
A block of mass 0.50 kg is moving with a speed of 2.00 ms–1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is- a)0.16 J
- b)1.00 J
- c)0.67 J
- d)0.34 J
Correct answer is option 'C'. Can you explain this answer?
A block of mass 0.50 kg is moving with a speed of 2.00 ms–1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
a)
0.16 J
b)
1.00 J
c)
0.67 J
d)
0.34 J
|
Geetika Mukherjee answered |
To answer the question, we need to know the rest of the information. Could you please provide the rest of the sentence?
A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is- a)20 m/s
- b)40 m/s
- c)10√30 m/s
- d)10 m/s
Correct answer is option 'B'. Can you explain this answer?
A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is
a)
20 m/s
b)
40 m/s
c)
10√30 m/s
d)
10 m/s
|
|
Maitri Basu answered |
To solve this problem, we can use the principle of conservation of energy.
At the top of the first hill, the ball has potential energy equal to its mass times the height of the hill:
PE1 = mgh1 = 20 kg * 9.8 m/s^2 * 100 m = 19600 J
When the ball reaches the bottom of the first hill, all of its potential energy is converted into kinetic energy:
KE1 = 19600 J
As the ball climbs up the second hill, it loses some of its kinetic energy to potential energy. The change in potential energy is equal to the change in kinetic energy:
PE2 = KE1 - KE2
At the top of the second hill, the ball has potential energy equal to its mass times the height of the hill:
PE2 = mgh2 = 20 kg * 9.8 m/s^2 * 30 m = 5880 J
The change in potential energy is therefore:
PE2 = KE1 - KE2
5880 J = 19600 J - KE2
KE2 = 13720 J
As the ball rolls down to the horizontal base, it loses potential energy and gains kinetic energy. The change in potential energy is equal to the change in kinetic energy:
PE3 = KE2 - KE3
At the base, the ball has potential energy equal to its mass times the height above the ground:
PE3 = mgh3 = 20 kg * 9.8 m/s^2 * 20 m = 3920 J
The change in potential energy is therefore:
PE3 = KE2 - KE3
3920 J = 13720 J - KE3
KE3 = 9800 J
Finally, the velocity attained by the ball can be found using the equation for kinetic energy:
KE = 1/2 * mv^2
Solving for velocity:
v = sqrt(2KE/m)
v = sqrt(2*9800 J / 20 kg)
v = sqrt(980 J) ≈ 31.3 m/s
Therefore, the velocity attained by the ball is approximately 31.3 m/s.
None of the provided answer choices (a) 20 m/s, b) 40 m/s, c) 10 m/s) are correct.
At the top of the first hill, the ball has potential energy equal to its mass times the height of the hill:
PE1 = mgh1 = 20 kg * 9.8 m/s^2 * 100 m = 19600 J
When the ball reaches the bottom of the first hill, all of its potential energy is converted into kinetic energy:
KE1 = 19600 J
As the ball climbs up the second hill, it loses some of its kinetic energy to potential energy. The change in potential energy is equal to the change in kinetic energy:
PE2 = KE1 - KE2
At the top of the second hill, the ball has potential energy equal to its mass times the height of the hill:
PE2 = mgh2 = 20 kg * 9.8 m/s^2 * 30 m = 5880 J
The change in potential energy is therefore:
PE2 = KE1 - KE2
5880 J = 19600 J - KE2
KE2 = 13720 J
As the ball rolls down to the horizontal base, it loses potential energy and gains kinetic energy. The change in potential energy is equal to the change in kinetic energy:
PE3 = KE2 - KE3
At the base, the ball has potential energy equal to its mass times the height above the ground:
PE3 = mgh3 = 20 kg * 9.8 m/s^2 * 20 m = 3920 J
The change in potential energy is therefore:
PE3 = KE2 - KE3
3920 J = 13720 J - KE3
KE3 = 9800 J
Finally, the velocity attained by the ball can be found using the equation for kinetic energy:
KE = 1/2 * mv^2
Solving for velocity:
v = sqrt(2KE/m)
v = sqrt(2*9800 J / 20 kg)
v = sqrt(980 J) ≈ 31.3 m/s
Therefore, the velocity attained by the ball is approximately 31.3 m/s.
None of the provided answer choices (a) 20 m/s, b) 40 m/s, c) 10 m/s) are correct.
This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2.
STATEMENT 1 : If stretched by the same amount work done on S1, Work done on S1 is more than S2 STATEMENT 2 : k1 < k2- a)Statement 1 is false, Statement 2 is true.
- b)Statement 1 is true, Statement 2 is false.
- c)Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1
- d)Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1
Correct answer is option 'A'. Can you explain this answer?
This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2.
STATEMENT 1 : If stretched by the same amount work done on S1, Work done on S1 is more than S2
If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2.
STATEMENT 1 : If stretched by the same amount work done on S1, Work done on S1 is more than S2
STATEMENT 2 : k1 < k2
a)
Statement 1 is false, Statement 2 is true.
b)
Statement 1 is true, Statement 2 is false.
c)
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1
d)
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1
|
Pragati Patel answered |
When force is same
.As W1 > W2
∴ k1 < k2
When extension is same
∴ W1 < W2
Statement 1 is false and statement 2 is true.
A 2 kg block slides on a horizontal floor with a speed of 4m/s.It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15N and spring constant is 10,000 N/m. The spring compresses by- a)8.5 cm
- b)5.5 cm
- c)2.5 cm
- d)11.0 cm
Correct answer is option 'B'. Can you explain this answer?
A 2 kg block slides on a horizontal floor with a speed of 4m/s.It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15N and spring constant is 10,000 N/m. The spring compresses by
a)
8.5 cm
b)
5.5 cm
c)
2.5 cm
d)
11.0 cm
|
Gauri Chauhan answered |
Let the block compress the spring by x before stopping. kinetic energy of the block = (P.E of compressed spring) + work done against friction.
10,000x2 + 30x – 32 = 0
⇒ 5000x2 + 15x - 16= 0
= 0.055m = 5.5cm.
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2 :- a) 9.89 × 10–3 kg
- b)12.89 × 10–3 kg
- c)2.45 × 10–3 kg
- d) 6.45 × 10–3 kg
Correct answer is option 'B'. Can you explain this answer?
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2 :
a)
9.89 × 10–3 kg
b)
12.89 × 10–3 kg
c)
2.45 × 10–3 kg
d)
6.45 × 10–3 kg
|
Nabanita Singh answered |
Consider the following two statements : A Linear momentum of a system of particles is zero B. Kinetic energy of a system of particles is zero.Then- a)A does not imply B and B does not imply A
- b)A implies B but B does not imply A
- c)A does not imply B but B implies A
- d)A implies B and B implies A
Correct answer is option 'C'. Can you explain this answer?
Consider the following two statements : A Linear momentum of a system of particles is zero B. Kinetic energy of a system of particles is zero.Then
a)
A does not imply B and B does not imply A
b)
A implies B but B does not imply A
c)
A does not imply B but B implies A
d)
A implies B and B implies A
|
Raksha Iyer answered |
The correct answer is option 'C': A does not imply B but B implies A.
Explanation:
To understand why option 'C' is the correct answer, let's analyze each statement individually.
Statement A: "Linear momentum of a system of particles is zero"
Linear momentum is defined as the product of mass and velocity. If the linear momentum of a system of particles is zero, it means that the total mass of the system is zero or the total velocity of the system is zero or both.
Statement B: "Kinetic energy of a system of particles is zero"
Kinetic energy is defined as the energy possessed by an object due to its motion. If the kinetic energy of a system of particles is zero, it means that the total energy of motion of all the particles in the system is zero.
Now let's analyze the relationship between these two statements.
1. A does not imply B:
If the linear momentum of a system of particles is zero, it does not necessarily mean that the kinetic energy of the system is zero. The kinetic energy depends not only on the motion but also on the masses of the particles. Even if the velocity is zero, the masses can still contribute to the kinetic energy.
2. B does not imply A:
If the kinetic energy of a system of particles is zero, it does not necessarily mean that the linear momentum of the system is zero. The kinetic energy can be zero if the particles are at rest or if they have equal and opposite velocities. However, the linear momentum can still be non-zero if the masses of the particles are not equal or if their velocities are not opposite.
3. B implies A:
If the kinetic energy of a system of particles is zero, it implies that the total velocity of the system is zero. This is because kinetic energy is directly proportional to the square of velocity. If the kinetic energy is zero, the velocity must also be zero.
Therefore, the correct answer is option 'C': A does not imply B but B implies A.
Explanation:
To understand why option 'C' is the correct answer, let's analyze each statement individually.
Statement A: "Linear momentum of a system of particles is zero"
Linear momentum is defined as the product of mass and velocity. If the linear momentum of a system of particles is zero, it means that the total mass of the system is zero or the total velocity of the system is zero or both.
Statement B: "Kinetic energy of a system of particles is zero"
Kinetic energy is defined as the energy possessed by an object due to its motion. If the kinetic energy of a system of particles is zero, it means that the total energy of motion of all the particles in the system is zero.
Now let's analyze the relationship between these two statements.
1. A does not imply B:
If the linear momentum of a system of particles is zero, it does not necessarily mean that the kinetic energy of the system is zero. The kinetic energy depends not only on the motion but also on the masses of the particles. Even if the velocity is zero, the masses can still contribute to the kinetic energy.
2. B does not imply A:
If the kinetic energy of a system of particles is zero, it does not necessarily mean that the linear momentum of the system is zero. The kinetic energy can be zero if the particles are at rest or if they have equal and opposite velocities. However, the linear momentum can still be non-zero if the masses of the particles are not equal or if their velocities are not opposite.
3. B implies A:
If the kinetic energy of a system of particles is zero, it implies that the total velocity of the system is zero. This is because kinetic energy is directly proportional to the square of velocity. If the kinetic energy is zero, the velocity must also be zero.
Therefore, the correct answer is option 'C': A does not imply B but B implies A.
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is - a)0.2 J
- b)10 J
- c)20 J
- d)0.1 J
Correct answer is option 'D'. Can you explain this answer?
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is
a)
0.2 J
b)
10 J
c)
20 J
d)
0.1 J
|
Soumya Nambiar answered |
Understanding Elastic Energy
When a wire is stretched by a weight, it stores elastic potential energy. The formula to calculate the elastic energy (U) stored in a stretched wire is given by:
U = (1/2) * F * x
Where:
- F is the force (weight) applied,
- x is the extension (stretch) of the wire.
Given Data
- Weight (F) = 200 N
- Extension (x) = 1 mm = 0.001 m (conversion to meters is necessary for calculations)
Calculating the Elastic Energy
1. Substituting Values:
- F = 200 N
- x = 0.001 m
2. Using the Formula:
U = (1/2) * 200 N * 0.001 m
U = (1/2) * 200 * 0.001
U = 0.1 J
Conclusion
The elastic energy stored in the wire is 0.1 Joules, which corresponds to option 'D'.
This calculation shows that the energy stored in the wire is relatively small, highlighting the efficiency of the material in storing potential energy under tension.
When a wire is stretched by a weight, it stores elastic potential energy. The formula to calculate the elastic energy (U) stored in a stretched wire is given by:
U = (1/2) * F * x
Where:
- F is the force (weight) applied,
- x is the extension (stretch) of the wire.
Given Data
- Weight (F) = 200 N
- Extension (x) = 1 mm = 0.001 m (conversion to meters is necessary for calculations)
Calculating the Elastic Energy
1. Substituting Values:
- F = 200 N
- x = 0.001 m
2. Using the Formula:
U = (1/2) * 200 N * 0.001 m
U = (1/2) * 200 * 0.001
U = 0.1 J
Conclusion
The elastic energy stored in the wire is 0.1 Joules, which corresponds to option 'D'.
This calculation shows that the energy stored in the wire is relatively small, highlighting the efficiency of the material in storing potential energy under tension.
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by 
where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is 
is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is is
where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is isa)
b)
c)
d)
|
|
Nabanita Singh answered |
At equilibrium : 
The potential energy of a 1 kg particle free to move along
the x-axis is given by
The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is- a)
- b) √2
- c)
- d)2
Correct answer is option 'A'. Can you explain this answer?
The potential energy of a 1 kg particle free to move along
the x-axis is given by
the x-axis is given by
The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is
a)
b)
√2
c)
d)
2
|
|
Nabanita Singh answered |
Velocity is maximum when K.E. is maximum For minimum. P.E.,
A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = –kx + ax3. Here k and a are positive constants. For x≥ 0 , the functional form of the potential energy U(x) of the particle is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = –kx + ax3. Here k and a are positive constants. For x≥ 0 , the functional form of the potential energy U(x) of the particle is
a)
b)
c)
d)
|
Mehul Chavan answered |
we have potential energy zero twice (out of which one is at origin).Also, when we put x = 0 in the given function,
should be zero. These characteristics are represented by (d).A body of mass ‘m’, accelerates uniformly from rest to ‘v1’ in time ‘t1’. The instantaneous power delivered to the body as a function of time ‘t’ is
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A body of mass ‘m’, accelerates uniformly from rest to ‘v1’ in time ‘t1’. The instantaneous power delivered to the body as a function of time ‘t’ is
a)
b)
c)
d)
|
|
Nabanita Singh answered |
To find the instantaneous power delivered to a body of mass mmm accelerating uniformly from rest to v1v_1v1 in time t1t_1t1, we first need to determine the expression for the velocity as a function of time and then use it to find the power.
-
Velocity as a function of time:Since the body starts from rest and accelerates uniformly, the velocity v(t)v(t)v(t) at any time ttt is given by:
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k2rt2 where k is a constant. The power delivered to the particles by the force acting on it is:- a)2π mk2r2 t
- b)mk2r2t
- c)
- d)zero
Correct answer is option 'B'. Can you explain this answer?
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k2rt2 where k is a constant. The power delivered to the particles by the force acting on it is:
a)
2π mk2r2 t
b)
mk2r2t
c)
d)
zero
|
|
Devika Banerjee answered |
The centripetal acceleration
A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is
a)
b)
c)
d)
|
|
Mehul Chavan answered |
Applying the principle of conservation of energy (K.E.)B + (P.E.)B = (K.E.)A + (P.E.)A,
we get
Change in velocity =
Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?
- a)4
- b)3
- c)2
- d)1
Correct answer is option 'C'. Can you explain this answer?
Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?
a)
4
b)
3
c)
2
d)
1
|
|
Devika Banerjee answered |
Let the radius of the circle be r. Then the two distance travelled by the two particles before first collision is 2πr. Therefore
2v ´ t + v ´ t =2pr
where t is the time taken for first collision to occur.
Distance travelled by particle with velocity v is equal to 
Therefore the collision occurs at B.
As the collision is elastic and the particles have equal masses, the velocities will interchange as shown in the figure. According to the same reasoning as above, the 2nd collision will take place at C and the velocities will again interchange.
With the same reasoning the 3rd collision will occur at the point A. Thus there will be two elastic collisions before the particles again reach at A.
With the same reasoning the 3rd collision will occur at the point A. Thus there will be two elastic collisions before the particles again reach at A.
A body of mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous power delivered to the body as a function of time is given by- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A body of mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous power delivered to the body as a function of time is given by
a)
b)
c)
d)
|
|
Pragati Patel answered |
u = 0; v = u + aT; v = aT
Instantaneous power = F × v = m. a. at = m.a2.t
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ‘t’ is proportional to- a)t3/4
- b)t3/2
- c)t1/4
- d)t1/2
Correct answer is option 'B'. Can you explain this answer?
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ‘t’ is proportional to
a)
t3/4
b)
t3/2
c)
t1/4
d)
t1/2
|
|
Nabanita Singh answered |
We know that F × v = Power
∴ F ´ v=c where c = constant
.
is applied over a particle which displaces it from its origin to the point 
m. The work done on the particle in joules is
A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2).
Q.2. The speed of the block at point C, immediately before it leaves the second incline is- a)√120 m/s
- b)√105 m/s
- c)√90 m/s
- d)√75 m/s
Correct answer is option 'B'. Can you explain this answer?
A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2).
Q.2. The speed of the block at point C, immediately before it leaves the second incline is
a)
√120 m/s
b)
√105 m/s
c)
√90 m/s
d)
√75 m/s
|
Pallavi Desai answered |
Applying mechanical energy conservation .Mechanical energy at B = Mechanical energy at C
A force F = - K 
(where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x axis to the point (a, 0), and then parallel to the y axis to the point (a, a), The total work done by the force F on the particle is- a)– 2Ka2
- b)2Ka2
- c)– Ka2
- d)Ka 2
Correct answer is option 'C'. Can you explain this answer?
A force F = - K (where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x axis to the point (a, 0), and then parallel to the y axis to the point (a, a), The total work done by the force F on the particle is
(where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x axis to the point (a, 0), and then parallel to the y axis to the point (a, a), The total work done by the force F on the particle isa)
– 2Ka2
b)
2Ka2
c)
– Ka2
d)
Ka 2
|
Pragati Nair answered |
The expression of work done by the variable force F on the particle is given by
In going from (0, 0) to (a, 0), the coordinate of x varies from 0 to 'a', while that of y remains zero. Hence, the work done along this path is :
In going from (a, 0) to (a, a) the coordinate of x remains constant (= a) while that of y changes from 0 to 'a'.
Hence, the work done along this path is
Hence, the work done along this path is
Hence, W = W1 + W2 = – ka2
The work done on a particle of mass m by a force,
(K being a constant of appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x – y plane is- a)
- b)
- c)
- d)0
Correct answer is option 'D'. Can you explain this answer?
The work done on a particle of mass m by a force,
(K being a constant of appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x – y plane is
a)
b)
c)
d)
0
|
Mira Roy answered |
Let us consider a point on the circle The equation of circle is x2 + y2 = a2 The force is
The force acts radially outwards as shown in the figure and the displacement is tangential to the circular path.
Therefore the angle between the force and displacement is 90° and W = 0 option (d) is correct.
Therefore the angle between the force and displacement is 90° and W = 0 option (d) is correct.
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface.
The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figure are only illustrative and not to the scale.- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface.
The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figure are only illustrative and not to the scale.
The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figure are only illustrative and not to the scale.
a)
b)
c)
d)
|
Ishita Reddy answered |
First the kinetic energy will increase as per eq (1). As the balls touches the ground it starts deforming and loses its K.E. (K.E. converting into elastic potential energy). When the deformation is maximum, K.E. = 0.
The ball then again regain its shape when its elastic potential energy changes into K.E. As the ball moves up it loses K.E. and gain gravitational potential energy.
These characteristics are according to graph (b).
The ball then again regain its shape when its elastic potential energy changes into K.E. As the ball moves up it loses K.E. and gain gravitational potential energy.
These characteristics are according to graph (b).
A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)
Q.5.The speed of the block when it reaches the point Q is- a)5 ms–1
- b)10 ms–1
- c)10√3 ms-1
- d)20ms–1
Correct answer is option 'B'. Can you explain this answer?
A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)
Q.5.The speed of the block when it reaches the point Q is
a)
5 ms–1
b)
10 ms–1
c)
10√3 ms-1
d)
20ms–1
|
|
Gauri Chauhan answered |
From (2), mv2 = 100 ∴ v = 10 ms–1
(b) is the correct option.
A Particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that- a)its kinetic energy is constant
- b)its acceleration is constant
- c)its velocity is constant
- d)it moves in a straight line
Correct answer is option 'A'. Can you explain this answer?
A Particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that
a)
its kinetic energy is constant
b)
its acceleration is constant
c)
its velocity is constant
d)
it moves in a straight line
|
Manisha Choudhary answered |
Work done by such force is always zero since force is acting in a direction perpendicular to velocity. ∴ from work-energy theorem = ΔK=0 K remains constant.
An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end.
The mass is released with the spring initially unstretched.
Then the maximum extension in the spring is - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end.
The mass is released with the spring initially unstretched.
Then the maximum extension in the spring is
The mass is released with the spring initially unstretched.
Then the maximum extension in the spring is
a)
b)
c)
d)
|
|
Devika Banerjee answered |
The above situation can also be looked upon as the decrease in the gravitational potential energy of spring mass system is equal to the gain in spring elastic potential energy.
If a machine is lubricated with oil (1980)- a)the mechanical advantage of the machine increases.
- b)the mechanical efficiency of the machine increases.
- c)both its mechanical advantage and efficiency increase.
- d)its efficiency increases, but its mechanical advantage decreases.
Correct answer is option 'B'. Can you explain this answer?
If a machine is lubricated with oil (1980)
a)
the mechanical advantage of the machine increases.
b)
the mechanical efficiency of the machine increases.
c)
both its mechanical advantage and efficiency increase.
d)
its efficiency increases, but its mechanical advantage decreases.
|
|
Ishita Reddy answered |
Mechanical efficiency =
The output work will increase because the friction becomes less. Thus the mechanical efficiency increases.
A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4k, respectively (see fig. I).
The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (figure II) and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. The ratio y/x is –
- a)4
- b)2
- c)1/2
- d)1/4
Correct answer is option 'C'. Can you explain this answer?
A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4k, respectively (see fig. I).
The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (figure II) and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. The ratio y/x is –
The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (figure II) and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. The ratio y/x is –
a)
4
b)
2
c)
1/2
d)
1/4
|
|
Dipika Choudhury answered |
When the block B is displaced towards wall 1, only spring S1 is compressed and S2 is in its natural state.
This happens because the other end of S2 is not attached to the wall but is free. Therefore the energy stored in the system = 1/2
k1 x2 . When the block is released, it will come back to the equilibrium position, gain momentum, overshoot to equilibrium position and move towards wall 2. As this happens, the spring S1 comes to its natural length and S2 gets compressed. As there are no frictional forces involved, the P.E. stored in the spring S1 gets stored as the P.E. of spring S2 when the block B reaches its extreme position after compressing S2 by y.
This happens because the other end of S2 is not attached to the wall but is free. Therefore the energy stored in the system = 1/2
k1 x2 . When the block is released, it will come back to the equilibrium position, gain momentum, overshoot to equilibrium position and move towards wall 2. As this happens, the spring S1 comes to its natural length and S2 gets compressed. As there are no frictional forces involved, the P.E. stored in the spring S1 gets stored as the P.E. of spring S2 when the block B reaches its extreme position after compressing S2 by y.
A small ball starts moving from A over a fixed track as shown in the figure. Surface AB has friction. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respectively.
Then
- a)
- b)
- c)
- d)
Correct answer is option 'A,B'. Can you explain this answer?
A small ball starts moving from A over a fixed track as shown in the figure. Surface AB has friction. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respectively.
Then
Then
a)
b)
c)
d)
|
|
Devika Banerjee answered |
At point A, potential energy of the ball = mghA
At point B, potential energy of the ball = 0
At point C, potential energy of the ball = mghC
Total energy at point A, EA = KA + mghA
Total energy at point B, EB = KB
Total energy at point C, EC = KC + mghC
According to the law of conservation of energy.
EA = EB = EC ... (i)
EA = EB ⇒ EC > KC ...(ii)
EA = EC
KA + mghA = KB + mghC
Option (b) is correct
From (i),(ii) and (iv), we get hA> hC; KB > KC.Option (a) is correct.
A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)
Q.4. The magnitude of the normal reaction that acts on the block at the point Q is - a)7.5N
- b)8.6N
- c)11.5N
- d)22.5N
Correct answer is option 'A'. Can you explain this answer?
A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)
Q.4. The magnitude of the normal reaction that acts on the block at the point Q is
a)
7.5N
b)
8.6N
c)
11.5N
d)
22.5N
|
|
Nabanita Singh answered |
Loss in P.E. = mg × 40 sin 30° = 200 J
Work done in over coming friction = 150 J
∴ K.E. possessed by the particle = 50 J
From (1) and (2) 
(a) is the correct option.
(a) is the correct option.A particle moves in a straight line with retar dation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to- a)x
- b)ex
- c)x2
- d)logex
Correct answer is option 'C'. Can you explain this answer?
A particle moves in a straight line with retar dation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to
a)
x
b)
ex
c)
x2
d)
logex
|
Advait Ghoshal answered |
Given : retardation µ displacement
i.e., a =-kx
∴ Loss in kinetic energy,∴ DKαx 2
A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is
a)
b)
c)
d)
|
|
Devika Banerjee answered |
The forces acting on the bead as seen by the observer in the accelerated frame are : (a) N ; (b) mg ; (c) ma (pseudo force).
Let θ is the angle which the tangent at P makes with the X- axis. As the bead is in equilibrium with respect to the wire, therefore N sinθ = ma and N cos θ = mg
Let θ is the angle which the tangent at P makes with the X- axis. As the bead is in equilibrium with respect to the wire, therefore N sinθ = ma and N cos θ = mg
From (i) & (ii)
A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2).
Q.3. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is –- a)√30 m/s
- b)√15 m/s
- c)0
- d)-√15 m/s
Correct answer is option 'C'. Can you explain this answer?
A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2).
Q.3. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is –
a)
√30 m/s
b)
√15 m/s
c)
0
d)
-√15 m/s
|
|
Advait Ghoshal answered |
The velocity of the block along BC just before collision is v cos 30°. The impact forces act perpendicular to the surface so the component of velocity along the incline remains unchanged.
Also since the collision is elastic, the vertical component of velocity (v sin 30°) before collision changes in direction, the magnitude remaining the same as shown in the figure. So the rectangular components of velocity after collision are as shown in the figure.
This means that the final velocity of the block should be horizontal making an angle 30º with BC. Therefore the vertical component of the final velocity of the block is zero.
This means that the final velocity of the block should be horizontal making an angle 30º with BC. Therefore the vertical component of the final velocity of the block is zero.
A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a time-dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is
- a)4.50 J
- b)7.50 J
- c)5.06 J
- d)14.06 J
Correct answer is option 'C'. Can you explain this answer?
A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a time-dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is
a)
4.50 J
b)
7.50 J
c)
5.06 J
d)
14.06 J
|
|
Amrutha Chaudhary answered |
Area under F – t graph gives the impulse or the change in the linear momentum of the body. As the initial velocity (and therefore the initial linear momentum) of the body is zero, the area under F – t graph gives the final linear momentum of the body.
Area of ΔAOB
Area of ΔAOB
A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2).
Q.1. The speed of the block at point B immediately after it strikes the second incline is –
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2).
Q.1. The speed of the block at point B immediately after it strikes the second incline is –
a)
b)
c)
d)
|
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Gauri Chauhan answered |
As the inclined plane is frictionless, The K. E . at B = P.E. at A
Δh = 3 m
This is the velocity of the block just before collision.
This velocity makes an angle of 30° with the vertical.
This velocity makes an angle of 30° with the vertical.
Also in right angled triangle BEC, ∠EBC= 60°.
Therefore v makes an angle of 30° with the second inclined plane BC. The component of v along BC is v cos 30°.
Therefore v makes an angle of 30° with the second inclined plane BC. The component of v along BC is v cos 30°.
It is given that the collision at B is perfectly inelastic therefore the impact forces act normal to the plane such that the vertical component of velocity becomes zero.
The component of velocity along the incline BC remains unchanged and is equal to v cos 30°
The component of velocity along the incline BC remains unchanged and is equal to v cos 30°
A particle is acted by a force F = kx, where k is a +ve constant. Its potential energy at x = 0 is zero. Which curve correctly represents the variation of potential energy of the block with respect t o x- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A particle is acted by a force F = kx, where k is a +ve constant. Its potential energy at x = 0 is zero. Which curve correctly represents the variation of potential energy of the block with respect t o x
a)
b)
c)
d)
|
|
Devika Banerjee answered |
We know that ΔU = – W for conservative forces
If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1,2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3
- a)W1> W2 > W3
- b)W1= W2 = W3
- c)W1< W2 < W3
- d) W2> W1 > W3
Correct answer is option 'B'. Can you explain this answer?
If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1,2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3
a)
W1> W2 > W3
b)
W1= W2 = W3
c)
W1< W2 < W3
d)
W2> W1 > W3
|
|
Ishita Reddy answered |
Note : In a conservative field work done does not depend on the path. The gravitational field is a conservative field.
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