All Exams >
JAMB >
Physics for JAMB >
All Questions
All questions of Work, Energy & Power for JAMB Exam
For what angle between Force and Displacement will the work done be positive?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For what angle between Force and Displacement will the work done be positive?
a)
b)
c)
d)
|
Om Desai answered |
- If a force acting on a body has a component in the opposite direction of displacement, the work done is negative.
- So when a body slides against a rough horizontal surface, its displacement is opposite to that of the force of friction. The work done by the friction is negative.
Work done by gravitational force on a man, in lifting a bucket out of the well by rope tied to the bucket is- a)negative
- b)positive
- c)zero
- d)infinity
Correct answer is option 'A'. Can you explain this answer?
Work done by gravitational force on a man, in lifting a bucket out of the well by rope tied to the bucket is
a)
negative
b)
positive
c)
zero
d)
infinity
|
|
Om Desai answered |
When the man pulls the bucket outside the well the gravitational potential of the man + bucket system increases and hence the work done by the gravitational force is negative.
An engine generates a power of 10 kW. In how much time will it raise a mass of 200 kg through a height of 40m.- a)4 sec
- b)10 sec
- c)5 sec
- d)8 sec
Correct answer is option 'D'. Can you explain this answer?
An engine generates a power of 10 kW. In how much time will it raise a mass of 200 kg through a height of 40m.
a)
4 sec
b)
10 sec
c)
5 sec
d)
8 sec
|
Mira Sharma answered |
Power = work done / time
work = force x distance
so power = (force x distance) / time
time = (force x distance) / power
= (2000 N x 40m) / 10000W = 8 seconds
A body dropped from height H reaches the ground with a speed of 
- a)mgh
- b)-0.28mgH
- c)2gh
- d)0.5gh
Correct answer is option 'B'. Can you explain this answer?
A body dropped from height H reaches the ground with a speed of
a)
mgh
b)
-0.28mgH
c)
2gh
d)
0.5gh
|
Krishna Iyer answered |
The forces acting on the body are the force of gravity and the air friction. By work energy theorem, the total work done on the body is
W=1/2m(1.2√gH)2−0=0.72mgH
The work done by the force of gravity is mgH.
. Hence, the work done by the air friction is,
=0.72mgH-mgH
=00.28mgH.
W=1/2m(1.2√gH)2−0=0.72mgH
The work done by the force of gravity is mgH.
. Hence, the work done by the air friction is,
=0.72mgH-mgH
=00.28mgH.
What is the dimension of k/m where k is the force constant and m is the mass of the oscillating object?- a)[T2]
- b)[T1]
- c)[T-1]
- d)[T-2]
Correct answer is option 'D'. Can you explain this answer?
What is the dimension of k/m where k is the force constant and m is the mass of the oscillating object?
a)
[T2]
b)
[T1]
c)
[T-1]
d)
[T-2]
|
Preeti Khanna answered |
F=k x (compression in string).
X=L.
a=LT-2.
F=ma.
ma= k L.
LT-2/L=k/m.
T-2=k/m.
X=L.
a=LT-2.
F=ma.
ma= k L.
LT-2/L=k/m.
T-2=k/m.
A person 'A' pushes a body of 5 kg placed on a rough surface of co-efficient of friction 0.3 by a distance of 4 m in 15 seconds. Another person 'B' pushes the body of 7 kg on same surface by a distance of 3 m in the same time. Who has more energy and power?- a)A
- b)B
- c)Both A and B have developed equal power but B has more energy
- d)Both A and B have developed equal power but A has more energy
Correct answer is option 'B'. Can you explain this answer?
A person 'A' pushes a body of 5 kg placed on a rough surface of co-efficient of friction 0.3 by a distance of 4 m in 15 seconds. Another person 'B' pushes the body of 7 kg on same surface by a distance of 3 m in the same time. Who has more energy and power?
a)
A
b)
B
c)
Both A and B have developed equal power but B has more energy
d)
Both A and B have developed equal power but A has more energy
|
Gaurav Kumar answered |
For person A,
Mass of body, m1 = 5kg
Coefficient of friction, μ =0.3
So, The force of friction,
F1= μm1g=0.3 X 5 X 10=15N
Displacement, S1=4m
Therefore,
Work done, W1=15X4=60J
Time taken, t1=15sec
So, Power, P1=60/15=4W
For Person B
Mass of body, m2=7kg
Coefficient of friction, μ =0.3
So, The force of friction,
F1= μm2g=0.3 X 7 X 10=21N
Displacement, S2=3m
Therefore,
Work done, W2=21X3=63J
Time taken, t2=15sec
So, Power, P3=63/15=4.2W
Mass of body, m1 = 5kg
Coefficient of friction, μ =0.3
So, The force of friction,
F1= μm1g=0.3 X 5 X 10=15N
Displacement, S1=4m
Therefore,
Work done, W1=15X4=60J
Time taken, t1=15sec
So, Power, P1=60/15=4W
For Person B
Mass of body, m2=7kg
Coefficient of friction, μ =0.3
So, The force of friction,
F1= μm2g=0.3 X 7 X 10=21N
Displacement, S2=3m
Therefore,
Work done, W2=21X3=63J
Time taken, t2=15sec
So, Power, P3=63/15=4.2W
So person B has more energy and power.
A man A of mass 80 kg runs up a staircase in 12 seconds. Another man of mass 60 kg runs up the same staircase in 11 seconds. The ratio of powers of A and B is :
- a)11:12
- b)9:11
- c)12:11
- d)11:9
Correct answer is option 'D'. Can you explain this answer?
A man A of mass 80 kg runs up a staircase in 12 seconds. Another man of mass 60 kg runs up the same staircase in 11 seconds. The ratio of powers of A and B is :
a)
11:12
b)
9:11
c)
12:11
d)
11:9
|
Naina Sharma answered |
According to question, 60 kg man runs up a staircase in 12 sec.
m₁ = 80 kg
t₁ = 12 sec
=> A 50 kg man runs up the same staircase in 11 sec:
m₂ = 60 kg
t₂ = 11 sec
=> Rate of work done is equals to power.
∴ P = E/ t and E = mgh
∴ P₁ = m₁gh / t₁
P₂ = m₂gh/ t₂
=> The ratio of Power or Rate of work done is:
P₁/P₂ = m₁gh / t₁ / m₂gh/ t₂
= m₁ * t₂ / m₂ * t₁
= 80x11/60x12
= 11 : 9
Thus, the ratio of rate of doing their work is 11 : 9.
An electric motor creates a tension of 4500 N in a hoisting cable and reels it in at the rate of 2 m/s. What is the power of the electric motor?
- a)225W
- b)15kW
- c)9kW
- d)900 W
Correct answer is option 'C'. Can you explain this answer?
An electric motor creates a tension of 4500 N in a hoisting cable and reels it in at the rate of 2 m/s. What is the power of the electric motor?
a)
225W
b)
15kW
c)
9kW
d)
900 W
|
|
Krishna Iyer answered |
F=4500N
v=2ms−1
The power of motor is given by
P=F.v
P=4500×2
P=9000W=9kW
v=2ms−1
The power of motor is given by
P=F.v
P=4500×2
P=9000W=9kW
You are in a lift moving from the 3rd floor to the 12th floor, having a height difference of H = 40 m. Your mass is M = 80 kg and the acceleration due to gravity is g = 10 m/s2. If the elevator moves at a constant speed without stopping, what is the total work performed by all of the forces together (the net force) acting on you?- a)Mgh
- b)-Mgh
- c)0
- d)Mg
Correct answer is option 'C'. Can you explain this answer?
You are in a lift moving from the 3rd floor to the 12th floor, having a height difference of H = 40 m. Your mass is M = 80 kg and the acceleration due to gravity is g = 10 m/s2. If the elevator moves at a constant speed without stopping, what is the total work performed by all of the forces together (the net force) acting on you?
a)
Mgh
b)
-Mgh
c)
0
d)
Mg
|
Adit Raj answered |
Net work done is equal to change in its kinetic energy
KE=1/2m(vf^2-vi^2)...since vf=vi so net work done is zero
KE=1/2m(vf^2-vi^2)...since vf=vi so net work done is zero
Time rate at which work is done by a force is- a)Power
- b)Torque
- c)Centrifugal Force
- d)Acceleration
Correct answer is option 'A'. Can you explain this answer?
Time rate at which work is done by a force is
a)
Power
b)
Torque
c)
Centrifugal Force
d)
Acceleration
|
|
Gaurav Kumar answered |
We know that Work done, W = F.s
where F is force and s is displacement due to that force.
Thus rate of work done is:
where F is force and s is displacement due to that force.
Thus rate of work done is:
dW/dt = d(F.s)/dt
So as F is constant we get,
dW/dt = F.d(s)/dt = F.v = P (Power)
dW/dt = F.d(s)/dt = F.v = P (Power)
By how much does kinetic energy increase if the momentum is increased by 20%?- a)55 %
- b)20 %
- c)44 %
- d)60 %
Correct answer is option 'C'. Can you explain this answer?
By how much does kinetic energy increase if the momentum is increased by 20%?
a)
55 %
b)
20 %
c)
44 %
d)
60 %
|
Hansa Sharma answered |
The kinetic energy is given by:
KE= p2/2m
So, ΔKE = 2pΔp / 2m = pΔp / m
ΔKE / KE = (pΔp/m) * (2m/p2) = 2Δp / p
Since the momentum p increases by 20%, so the final momentum becomes 1.2p.
Hence, KEfinal = (1.2p)2 / 2m = 1.44p2 / 2m = 1.44KE
So, % change in KE = 44%
ΔKE / KE = (pΔp/m) * (2m/p2) = 2Δp / p
Since the momentum p increases by 20%, so the final momentum becomes 1.2p.
Hence, KEfinal = (1.2p)2 / 2m = 1.44p2 / 2m = 1.44KE
So, % change in KE = 44%
Output of a truck is 4500 J and its efficiency is 50%, the input energy provided to the truck is- a)5000 J
- b)900 J
- c)9000 J
- d)500 J
Correct answer is option 'C'. Can you explain this answer?
Output of a truck is 4500 J and its efficiency is 50%, the input energy provided to the truck is
a)
5000 J
b)
900 J
c)
9000 J
d)
500 J
|
|
Naina Sharma answered |
► η = work output / heat input
► η = 50% = 50/100 = 1/2
► 1/2 = 4500 / Heat input
► Heat input = 9000 J
► η = 50% = 50/100 = 1/2
► 1/2 = 4500 / Heat input
► Heat input = 9000 J
There are two bodies X and Y with equal kinetic energy but different masses m and 4m respectively. The ratio of their linear momentum is-- a)1:2
- b)4:1
- c)1:√2
- d)1:4
Correct answer is option 'A'. Can you explain this answer?
There are two bodies X and Y with equal kinetic energy but different masses m and 4m respectively. The ratio of their linear momentum is-
a)
1:2
b)
4:1
c)
1:√2
d)
1:4
|
Suresh Iyer answered |
X and Y have equal kinetic energy but their masses are m and 4m respectively.
► 1/2 m1v12 = 1/2 m2v22
► mv12 = 4m * v22
► v1 : v2 = 2 : 1
Hence the ratio of their linear momentum is:
m1v1 : m2v2 = m * 2v : 4m * v = 1 : 2
► 1/2 m1v12 = 1/2 m2v22
► mv12 = 4m * v22
► v1 : v2 = 2 : 1
Hence the ratio of their linear momentum is:
m1v1 : m2v2 = m * 2v : 4m * v = 1 : 2
A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radian per second after 5 second. The total angular displacement is :- a)800 Red
- b)400 Red
- c)200 Red
- d)100 Red
Correct answer is option 'C'. Can you explain this answer?
A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radian per second after 5 second. The total angular displacement is :
a)
800 Red
b)
400 Red
c)
200 Red
d)
100 Red
|
Arjun Singhania answered |
As stated
if the angular velocity increases uniformly . The angular acceleration must be constant.
Hence the formula
s= ut+ 0.5 a t^2
Where.
u= initial angular velocity
a= angular acceleration (i.e = 80/5=16 rad/s^2)
S= angular displacement
t= time taken
Now.
s= 0 + 0.5 X 16*5*5 = 200
Hence angar displacement = 200 radians
A body moves under a force such that momentum of the body at any instant 
. What is the rate of work done by the force?- a)
- b)
- c)
- d)zero
Correct answer is option 'D'. Can you explain this answer?
A body moves under a force such that momentum of the body at any instant . What is the rate of work done by the force?
. What is the rate of work done by the force?a)
b)
c)
d)
zero
|
|
Hansa Sharma answered |
P = p Cos(t) i + p Sin(t) j
dP/dt = -p Sin(t) j + p Cos(t) i
Work done, W = P2/2m
Rate of work done = d/dt(P2/2m)
= 2P/2m x dP/dt
= P/m x dP/dt
= -p2.Sin(t).Cos(t) + p2.Sin(t).Cos(t)
=> Rate of work done = 0
Calculate the power developed when the force of 10 N is exerted on a body moving with velocity of 3.65 m/s.- a)36.5W
- b)3.65W
- c)365W
- d)0.365W
Correct answer is option 'A'. Can you explain this answer?
Calculate the power developed when the force of 10 N is exerted on a body moving with velocity of 3.65 m/s.
a)
36.5W
b)
3.65W
c)
365W
d)
0.365W
|
Lavanya Menon answered |
Power =F.V
i.e, Power = force x velocity.
here, force (f)= 10N
and velocity (v) = 3.65 m/s
so, Power = f x v,
i.e, 10 x 3.65 = 36.5 w.
i.e, Power = force x velocity.
here, force (f)= 10N
and velocity (v) = 3.65 m/s
so, Power = f x v,
i.e, 10 x 3.65 = 36.5 w.
How much work is done by a man weighing 80 kgf in raising a stone of weight 20 kgf to the top of a building of height 40 m? (g = 9.8 m/s2)- a)30,900 J
- b)29,678 J
- c)39,200 J
- d)25,000 J
Correct answer is option 'C'. Can you explain this answer?
How much work is done by a man weighing 80 kgf in raising a stone of weight 20 kgf to the top of a building of height 40 m? (g = 9.8 m/s2)
a)
30,900 J
b)
29,678 J
c)
39,200 J
d)
25,000 J
|
Sushil Kumar answered |
Total mass=80+20=100
w=mgh(where m=100, g=9.8, h=40m)
w=100×9.8×40
w=39200J
work done by man weighing 80 kg raising a stone of 20kgf to the top of building of Height is 40m is 39,200J
w=mgh(where m=100, g=9.8, h=40m)
w=100×9.8×40
w=39200J
work done by man weighing 80 kg raising a stone of 20kgf to the top of building of Height is 40m is 39,200J
An elastic spring of force constant k is compressed by an amount x. Then its potential energy will be- a)kx
- b)-kx
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
An elastic spring of force constant k is compressed by an amount x. Then its potential energy will be
a)
kx
b)
-kx
c)
d)
|
|
Anjana Sharma answered |
F= kx
Now as we know
f.ds = potential energy
differentiate
potential energy = kx .ds
k is a constant .... and diff of x .ds is = x^2 / 2
so here potential energy will be = 1/2 k x^2
How should the force applied on a body be varied with velocity to keep the power of force constant?- a)Force should be inversely proportional to the square root of the velocity of the body
- b)Force should be inversely proportional to the velocity of the body
- c)Force should be directly proportional to the velocity of the body
- d)Force should not be varied. It should remain constant with the velocity
Correct answer is option 'B'. Can you explain this answer?
How should the force applied on a body be varied with velocity to keep the power of force constant?
a)
Force should be inversely proportional to the square root of the velocity of the body
b)
Force should be inversely proportional to the velocity of the body
c)
Force should be directly proportional to the velocity of the body
d)
Force should not be varied. It should remain constant with the velocity
|
|
Anjana Sharma answered |
We know that power (P) = force (F) x velocity (v).
Since the body is moving with acceleration,v changes and as a result of that P also changes,F being constant.
For P to be constant, F v = constant or F ∝ 1/v i.e
Force should be inversely proportional to the velocity of the body.
Thus,as v increases, F should decrease to keep P constant.
In the given figure, what is the work done by the spring force?
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
In the given figure, what is the work done by the spring force?
a)
b)
c)
d)
|
|
Neha Joshi answered |
We know that for a spring with spring constant k, work done by the spring when a compression or expansion is done of some length say x is ½ kx2
Also as the work done is in the opposite direction to that of force applied, we get the sign of the force to be negative.
Also as the work done is in the opposite direction to that of force applied, we get the sign of the force to be negative.
A machine gun fires 60 bullets per minute, with a velocity of 700 m/s. If each bullet has a mass of 50g, find the power developed by the gun.
- a)1225 W
- b)12250 W
- c)122.5 W
- d)122 W
Correct answer is option 'B'. Can you explain this answer?
A machine gun fires 60 bullets per minute, with a velocity of 700 m/s. If each bullet has a mass of 50g, find the power developed by the gun.
a)
1225 W
b)
12250 W
c)
122.5 W
d)
122 W
|
Tejas Verma answered |
Each bullet will have KE = 1/2 * 0.05 * 700 * 700 = 12250 J
So for 60 bullets the energy given by machine in 60 second = 60 x 12250 J
Hence power developed = Energy / time =60 x12250/60 =12250W
Or power = 12.250 kW.
So for 60 bullets the energy given by machine in 60 second = 60 x 12250 J
Hence power developed = Energy / time =60 x12250/60 =12250W
Or power = 12.250 kW.
A vehicle of 50 quintal climbs up a hill of 10 m. The potential energy gained by it is (take g = 10 m/s2)- a)5 J
- b)5000 J
- c)500 J
- d)5 x 105 J
Correct answer is option 'D'. Can you explain this answer?
A vehicle of 50 quintal climbs up a hill of 10 m. The potential energy gained by it is (take g = 10 m/s2)
a)
5 J
b)
5000 J
c)
500 J
d)
5 x 105 J
|
|
Geetika Shah answered |
mass = 50 x 100 = 5000 kg, g = 10 m/s2, h = 10 m
Potential energy gained = mgh = 5000 x 10 x 10 = 5 x 105 J
Potential energy gained = mgh = 5000 x 10 x 10 = 5 x 105 J
Which of the following is not conserved in inelastic collision?- a)momentum
- b)kinetic energy
- c)both momentum and kinetic energy
- d)neither momentum nor kinetic energ
Correct answer is option 'B'. Can you explain this answer?
Which of the following is not conserved in inelastic collision?
a)
momentum
b)
kinetic energy
c)
both momentum and kinetic energy
d)
neither momentum nor kinetic energ
|
Rajesh Gupta answered |
In an inelastic collision, the force of reformation is not equal to the force of deformation and thus some amount of energy is lost. But still as no external force acts upon the system momentum is still conserved.
Which of the following statement is not related to conservative force?- a)Work done in closed path is zero
- b)Work done is recoverable
- c)Path independent
- d)Path dependent
Correct answer is option 'D'. Can you explain this answer?
Which of the following statement is not related to conservative force?
a)
Work done in closed path is zero
b)
Work done is recoverable
c)
Path independent
d)
Path dependent
|
|
Krishna Iyer answered |
A force is said to be conservative if the work done by or against the force on a body is independent of path followed by the body and depends only on initial and final positions.
Work done by or against the conservative force in moving a particle along a closed path is zero.
Work done by or against the conservative force in moving a particle along a closed path is zero.
Which of the following is not a unit of energy?- a)newton meter
- b)electron volt
- c)joule/meter
- d)kilowatt hour
Correct answer is option 'C'. Can you explain this answer?
Which of the following is not a unit of energy?
a)
newton meter
b)
electron volt
c)
joule/meter
d)
kilowatt hour
|
Gargey Dudhe answered |
C is correct option because energy The joule ( symbol: J) is a derived unit of energy in the International System of Units. It is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of its motion through a distance of one metre (1 newton metreor N⋅m). pls upvote and follow me.
When does the potential energy of a spring increase?- a)only when spring is stretched
- b)only when spring is compressed
- c)when spring is neither stretched nor compressed
- d)when spring is compressed or stretched
Correct answer is option 'D'. Can you explain this answer?
When does the potential energy of a spring increase?
a)
only when spring is stretched
b)
only when spring is compressed
c)
when spring is neither stretched nor compressed
d)
when spring is compressed or stretched
|
|
Gaurav Kumar answered |
Potential energy of a spring is proportional to the square of the difference of the springs length and its original length, hence whether it is compressed or stretched the potential the potential energy will eventually increase only.
Two moving particle P and Q are 10 m apart at a certain instant. The velocity of P is 8m/s making an angle 30° with the line joining P and Q and that of Q is 6m/s making an angle 30° with PQ as shown in the figure. Then angular velocity of P with respect to Q is 
- a)Zero
- b)0.1 rad/sec
- c)0.4 rad/sec
- d)0.7 rad sec
Correct answer is option 'D'. Can you explain this answer?
Two moving particle P and Q are 10 m apart at a certain instant. The velocity of P is 8m/s making an angle 30° with the line joining P and Q and that of Q is 6m/s making an angle 30° with PQ as shown in the figure. Then angular velocity of P with respect to Q is
a)
Zero
b)
0.1 rad/sec
c)
0.4 rad/sec
d)
0.7 rad sec
|
|
Krishna Iyer answered |
A machine, in an amusement park, consists of a cage at the end of one arm, hinged at O. The cage revolves along a vertical circle of radius r (ABCDEFGH) about its hinge O, at constant linear speedv = 
. The cage is so attached that the man of weight `w' standing on a weighing machine, inside the cage, is always vertical. Then which of the following is correct
- a)The reading of his weight on the machine is the same at all positions
- b)The weight reading at A is greater than the weight reading at E by 2 w.
- c)The weight reading at G = w
- d)The ratio of the weight reading at E to that at A=0
- e)The ratio of the weight reading at A to that at C=2
Correct answer is option 'B,C,D,E'. Can you explain this answer?
A machine, in an amusement park, consists of a cage at the end of one arm, hinged at O. The cage revolves along a vertical circle of radius r (ABCDEFGH) about its hinge O, at constant linear speed
v = . The cage is so attached that the man of weight `w' standing on a weighing machine, inside the cage, is always vertical. Then which of the following is correct
. The cage is so attached that the man of weight `w' standing on a weighing machine, inside the cage, is always vertical. Then which of the following is correcta)
The reading of his weight on the machine is the same at all positions
b)
The weight reading at A is greater than the weight reading at E by 2 w.
c)
The weight reading at G = w
d)
The ratio of the weight reading at E to that at A=0
e)
The ratio of the weight reading at A to that at C=2
|
Crafty Classes answered |
No work is done by a force on an object if- a)The force is always perpendicular to its velocity
- b)The force is always perpendicular to its acceleration
- c)The object is stationary but the point of application of the force moves on the object.
- d)The object moves in such a way that the point of application of the force remains fixed.
Correct answer is option 'A,B'. Can you explain this answer?
No work is done by a force on an object if
a)
The force is always perpendicular to its velocity
b)
The force is always perpendicular to its acceleration
c)
The object is stationary but the point of application of the force moves on the object.
d)
The object moves in such a way that the point of application of the force remains fixed.
|
|
Krishna Iyer answered |
Work done = f.ds
So on perpendicular application for product becomes zero so no work is done
So on perpendicular application for product becomes zero so no work is done
The potential energy of a body at height h is mgh. Then its kinetic energy just before hitting the ground is- a)mgh
- b)
- c)mv2
- d)
Correct answer is option 'A'. Can you explain this answer?
The potential energy of a body at height h is mgh. Then its kinetic energy just before hitting the ground is
a)
mgh
b)
c)
mv2
d)
|
|
Om Desai answered |
From the law of conservation of mechanical energy, the potential energy of a body at height h is converted into kinetic energy when it falls down. So, the magnitude of kinetic energy just before hitting the ground is equal to the potential energy at height h which is mgh.
Frictional force is an example of- a)ectrostatic force
- b)Non conservative force
- c)conservative force
- d)nuclear force
Correct answer is option 'B'. Can you explain this answer?
Frictional force is an example of
a)
ectrostatic force
b)
Non conservative force
c)
conservative force
d)
nuclear force
|
|
Om Desai answered |
The friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. This friction is applied tangentially to the surfaces in contact. Thus the main thing is that the forces on both of the surfaces act tangential to each other.
When conservative force does positive work on a body, the potential energy of the body:- a)increases
- b)decreases
- c)remains unaltered
- d)there is no relationship between force and potential energy
Correct answer is option 'B'. Can you explain this answer?
When conservative force does positive work on a body, the potential energy of the body:
a)
increases
b)
decreases
c)
remains unaltered
d)
there is no relationship between force and potential energy
|
|
Suresh Iyer answered |
We have a relation between potential energy and work done: 
The negative sign just shows the inverse relation between potential energy and work done by conservative force.
The negative sign just shows the inverse relation between potential energy and work done by conservative force.
A rigid body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is 
and is directed towards the centre. What is the work done by this force in moving the body over half the cirumference of the circle.- a)
- b) Zero
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A rigid body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is and is directed towards the centre. What is the work done by this force in moving the body over half the cirumference of the circle.
and is directed towards the centre. What is the work done by this force in moving the body over half the cirumference of the circle.a)
b)
Zero
c)
d)
|
|
Lavanya Menon answered |
Force = mv2/r ( towards the centre )
displacement = 2r
Work done = FScosѲ
here ,
= mv2/r × s × cos 90
= 0 answer.
From a waterfall, water is falling at the rate of 100 kg/s on the blades of a turbine. If the height of fall is 100 m, the power delivered to the turbine is- a)100kW
- b)100W
- c)1kW
- d)10kW
Correct answer is option 'A'. Can you explain this answer?
From a waterfall, water is falling at the rate of 100 kg/s on the blades of a turbine. If the height of fall is 100 m, the power delivered to the turbine is
a)
100kW
b)
100W
c)
1kW
d)
10kW
|
|
Nandini Patel answered |
Power = work done / time taken
or P = mgh / t
P = (m/t) gh = 100 x 10 x 100 = 100 kW
Which of the following statements is false:- a)kinetic energy is positiv
- b)potential energy is positive
- c)kinetic energy is negative
- d)potential energy is negative
Correct answer is option 'C'. Can you explain this answer?
Which of the following statements is false:
a)
kinetic energy is positiv
b)
potential energy is positive
c)
kinetic energy is negative
d)
potential energy is negative
|
Aiims answered |
K.E never negative as velocity negative then its square becomes it positive.
Select the odd one out- a)Viscous force
- b)Frictional force
- c)Electrostatic force
- d)Air-resistance
Correct answer is option 'C'. Can you explain this answer?
Select the odd one out
a)
Viscous force
b)
Frictional force
c)
Electrostatic force
d)
Air-resistance
|
|
Anjali Iyer answered |
The correct answer is muscular force because all these forces are non-contact forces and muscular force is a contact force.
Which of the following force is non-conservative- a)Restoring force of spring
- b)Force between two stationary masses
- c)Force between two stationary charges
- d)Human push or pull
Correct answer is option 'D'. Can you explain this answer?
Which of the following force is non-conservative
a)
Restoring force of spring
b)
Force between two stationary masses
c)
Force between two stationary charges
d)
Human push or pull
|
|
Anjali Iyer answered |
You mean all force that humans apply, or any push/pull force one person does. A conservative force is the property that the work done in moving a particle between two points is independent of the taken path. A conservative force is dependent only on the position of the object. Eg gravity.
For a conservative force, F is equal to- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For a conservative force, F is equal to
a)
b)
c)
d)
|
|
Pooja Shah answered |
Potential energy at a point is defined in terms the amount of work done, which is defined in terms of force and thus we get potential as an integral of force applied over some position x. Hence F(x) = -dU(x) / dx which is the differentiation of potential energy wrt position.
If a force acts perpendicular to the direction of motion of a body, what is the amount of work done?- a)Infinity
- b)Constant
- c)Zero
- d)sinθ
Correct answer is option 'C'. Can you explain this answer?
If a force acts perpendicular to the direction of motion of a body, what is the amount of work done?
a)
Infinity
b)
Constant
c)
Zero
d)
sinθ
|
|
Anjali Sharma answered |
If a force acts perpendicular to the direction of a body, the amount of work done is zero because there is no displacement in the direction of a force.
For a varying force work done can be expressed as a- a)product of force and displacement
- b)force divided by the displacement
- c)definite integral of force over displacement
- d)differentiation of force with displacement
Correct answer is option 'C'. Can you explain this answer?
For a varying force work done can be expressed as a
a)
product of force and displacement
b)
force divided by the displacement
c)
definite integral of force over displacement
d)
differentiation of force with displacement
|
|
Krishna Iyer answered |
For a variable force, the work done can be expressed as a definite integral of force over displacement for any system.In case of a variable force, the is calculated with the help of integration. For example, in the case of a spring, the force acting upon any object attached to a horizontal spring can be given as Fs = -kx, where k is the spring constant and x is the displacement of the object attached. We can see that this force is proportional to the displacement of the object from the equilibrium position, hence the force acting at each instant during the compression and extension of the spring will be different. Thus, the infinitesimally small contributions of work done during each instant are to be counted in order to calculate the total work done.
If momentum of an object is doubled, how does its kinetic energy change?- a)increases 4 times
- b)increases 2 times
- c)decreases 4 times
- d)decreases 2 times
Correct answer is option 'A'. Can you explain this answer?
If momentum of an object is doubled, how does its kinetic energy change?
a)
increases 4 times
b)
increases 2 times
c)
decreases 4 times
d)
decreases 2 times
|
|
Naina Bansal answered |
kinetic energy of object = 1/2 mv2
when the momentum of object is doubled,
then, kinetic energy of object = 1/2 m(2v)2
= 1/52 4mv2
= 4[1/2mv2]
so the momentum of an object after doubled is 4 times graeter.
A body is moved along a straight line path by a machine delivering constant power. The distance moved by the body in time t is proportional to:- a)t3/2
- b)t3/4
- c)t1/4
- d)t1/2
Correct answer is option 'A'. Can you explain this answer?
A body is moved along a straight line path by a machine delivering constant power. The distance moved by the body in time t is proportional to:
a)
t3/2
b)
t3/4
c)
t1/4
d)
t1/2
|
Shruti Mehta answered |
A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time 't' is proportional to.
Can you explain the answer of this question below:An astronomer has total weight of 80 kg. On reaching moon, he climbs up 10 m to enter his aircraft by the ladder. His gravitational potential energy is increased by 1000 J. The value of the gravitational acceleration on the moon is
- A:
2.22 kg
- B:
2.25 kg
- C:
2 N/kg
- D:
1.25 N/kg
The answer is d.
An astronomer has total weight of 80 kg. On reaching moon, he climbs up 10 m to enter his aircraft by the ladder. His gravitational potential energy is increased by 1000 J. The value of the gravitational acceleration on the moon is
2.22 kg
2.25 kg
2 N/kg
1.25 N/kg
|
aakash answered |
Mgh = 80 x 10 x g = 1000
g= 1000 / 800
g = 125 N/ Kg
g= 1000 / 800
g = 125 N/ Kg
A box of mass m is released from rest at position on the frictionless curved track shown. It slides a distance d along the track in time t to reach position 2, dropping a vertical distance h. Let v and a be the instantaneous speed and instantaneous acceleration, respectively, of the box at position 2. Which of the following equations is valid for this situation? 
- a)h = vt
- b)h = (1/2)gt2
- c)mgh = (1/2)mv2
- d)none
Correct answer is option 'C'. Can you explain this answer?
A box of mass m is released from rest at position on the frictionless curved track shown. It slides a distance d along the track in time t to reach position 2, dropping a vertical distance h. Let v and a be the instantaneous speed and instantaneous acceleration, respectively, of the box at position 2. Which of the following equations is valid for this situation?
a)
h = vt
b)
h = (1/2)gt2
c)
mgh = (1/2)mv2
d)
none
|
Imk Pathsala answered |
According to the Work Energy Theorem,
Change in PE = Change in KE
So, mgh = ½ mv2
Hence C
Forum id- 1775160
For complete circle v = √(5gl)
mgh = ½mv2
So, h = 2.5R
Change in PE = Change in KE
So, mgh = ½ mv2
Hence C
Forum id- 1775160
For complete circle v = √(5gl)
mgh = ½mv2
So, h = 2.5R
A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal smooth surface. The maximum tension the string can withstand is 16 newton. The maximum speed of revolution of the stone without breaking it, will be :- a)20 ms-1
- b)16 ms-1
- c)14 ms-1
- d)12 ms-1
Correct answer is option 'A,C,D'. Can you explain this answer?
A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal smooth surface. The maximum tension the string can withstand is 16 newton. The maximum speed of revolution of the stone without breaking it, will be :
a)
20 ms-1
b)
16 ms-1
c)
14 ms-1
d)
12 ms-1
|
|
Pooja Shah answered |
T= (mv2max)/r
16 = (16 v2max)/144
v2max = 12 m/s
16 = (16 v2max)/144
v2max = 12 m/s
The minimum value of H required so that the particle makes a complete vertical circle is given by- a)5 R
- b)4 R
- c)2.5 R
- d)2 R
Correct answer is option 'C'. Can you explain this answer?
The minimum value of H required so that the particle makes a complete vertical circle is given by
a)
5 R
b)
4 R
c)
2.5 R
d)
2 R
|
|
Hansa Sharma answered |
For complete circle v = √(5gl)
mgh = ½mv2
So, h = 2.5R
mgh = ½mv2
So, h = 2.5R
A wheel is subjected to uniform angular a cceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle q1 ; in the next 2 sec, it rotates through an additional angle q2. The ratio of q2 / q1 is-- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
A wheel is subjected to uniform angular a cceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle q1 ; in the next 2 sec, it rotates through an additional angle q2. The ratio of q2 / q1 is-
a)
1
b)
2
c)
3
d)
4
|
|
Mayank Chatterjee answered |
From the second equation of motion we get that
Q1 = 0 + ½ a x 4
Where a is constant angular acceleration
Thus Q1 = 2a
Similarly Q2 = 0 + ½ x a x 16
Thus Q2 = 8a
We get Q2 / Q1 = 4
Q1 = 0 + ½ a x 4
Where a is constant angular acceleration
Thus Q1 = 2a
Similarly Q2 = 0 + ½ x a x 16
Thus Q2 = 8a
We get Q2 / Q1 = 4
Chapter doubts & questions for Work, Energy & Power - Physics for JAMB 2025 is part of JAMB exam preparation. The chapters have been prepared according to the JAMB exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JAMB 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Work, Energy & Power - Physics for JAMB in English & Hindi are available as part of JAMB exam.
Download more important topics, notes, lectures and mock test series for JAMB Exam by signing up for free.
Physics for JAMB
259 videos|253 docs|230 tests
|
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup