All questions of Work , Energy and Power for NEET Exam

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.

K.E never negative as velocity negative then its square becomes it positive.

The odd one out among the given options is option 'C', which represents the Electrostatic force.

**Viscous Force:**
- Viscous force refers to the resistance that a fluid (liquid or gas) exerts on an object moving through it.
- It is a type of frictional force that opposes the relative motion of the object and the fluid.
- Viscous force is responsible for phenomena like drag experienced by objects moving through a fluid medium.
- Examples of viscous force include the resistance experienced by a car moving through air or a boat moving through water.

**Frictional Force:**
- Frictional force is the force that opposes the motion or attempted motion of an object past another object with which it is in contact.
- It arises due to the roughness or irregularities present on the surfaces in contact.
- Frictional force plays a crucial role in our daily lives, such as walking, driving a car, or holding objects.
- It can be both advantageous (e.g., walking) and disadvantageous (e.g., wearing out of machine parts).

**Electrostatic Force:**
- Electrostatic force is the force of attraction or repulsion between charged objects.
- It arises due to the electric charge carried by the objects.
- Objects with the same charges repel each other, while objects with opposite charges attract each other.
- Electrostatic force is responsible for various phenomena such as the attraction of clothes after being dried in a dryer or the repulsion of two magnets.

**Air-Resistance:**
- Air resistance, also known as drag, is a force that opposes the motion of an object through the air.
- It is caused by the collision of air molecules with the object in motion.
- Air resistance depends on factors such as the shape and size of the object, the speed of motion, and the density of the air.
- It affects objects moving through the air, like a falling parachute or a moving car.

**Explanation:**
The odd one out is electrostatic force because it is the only option that does not involve the interaction between objects in motion or with a fluid medium. Electrostatic force is related to the interaction between charged objects, whereas the other three forces (viscous force, frictional force, and air resistance) are associated with the motion of objects through a medium (fluid or air).

Kinetic energy is conserved in elastic collisions, whereas kinetic energy is converted into other forms of energy during an inelastic collision. In both types of collisions, momentum is conserved.

**Answer:**

The power exerted by the gravitational force can be calculated using the formula:

Power = force x velocity

When a body is projected vertically upwards, the only force acting on it is the gravitational force. The gravitational force is given by:

F = mg

Where:
F = gravitational force
m = mass of the body
g = acceleration due to gravity

Since the body reaches a height equal to the Earth's radius, the distance traveled by the body is 2 times the Earth's radius. Let's assume the Earth's radius is R.

Therefore, the work done against the gravitational force is given by:

Work = force x distance
= mg x 2R
= 2mgR

Now, the time taken to reach the maximum height is given by:

t = (2u sinθ) / g

Where:
t = time taken
u = initial velocity
θ = angle of projection
g = acceleration due to gravity

Since the body is projected vertically upwards, the angle of projection is 90° and the sine of 90° is 1.

Therefore, the time taken to reach the maximum height is:

t = (2u) / g

Now, the power exerted by the gravitational force can be calculated using the formula:

Power = Work / time
= (2mgR) / [(2u) / g]
= mg^2R/u

From this equation, we can see that the power exerted by the gravitational force is inversely proportional to the initial velocity (u). As the body reaches the highest position, the velocity becomes zero. Therefore, the power exerted by the gravitational force is greatest at the highest position of the body.

Hence, the correct answer is option 'A' - at the highest position of the body.

$^{-1}$ and 2 kg second part moving with a velocity of 8 ms$^{-1}$. What is the mass and velocity of the third part?

To solve this problem, we need to use the law of conservation of momentum. According to this law, the total momentum of a system of objects remains constant if there are no external forces acting on the system. In other words, the sum of the momenta of all the objects before the explosion is equal to the sum of the momenta of all the objects after the explosion.

Before the explosion, the rock had zero velocity, so its momentum was zero. After the explosion, the two parts that went off at right angles to each other have the following momenta:

First part: momentum = mass x velocity = 1 kg x 12 ms$^{-1}$ = 12 kg ms$^{-1}$

Second part: momentum = mass x velocity = 2 kg x 8 ms$^{-1}$ = 16 kg ms$^{-1}$

The total momentum of these two parts is:

Total momentum = 12 kg ms$^{-1}$ + 16 kg ms$^{-1}$ = 28 kg ms$^{-1}$

According to the law of conservation of momentum, the momentum of the third part must be equal and opposite to the total momentum of the first two parts. Let's call the mass of the third part "m" and its velocity "v". Then we have:

Momentum of third part = -28 kg ms$^{-1}$

Momentum = mass x velocity

Therefore:

-mv = -28 kg ms$^{-1}$

Solving for "m", we get:

m = 28/v

Now we can use the law of conservation of energy to find the velocity of the third part. According to this law, the total kinetic energy of a system of objects remains constant if there are no external forces acting on the system. In other words, the sum of the kinetic energies of all the objects before the explosion is equal to the sum of the kinetic energies of all the objects after the explosion.

Before the explosion, the rock had zero kinetic energy, so its total kinetic energy was zero. After the explosion, the two parts that went off at right angles to each other have the following kinetic energies:

First part: KE = 0.5 x mass x velocity$^2$ = 0.5 x 1 kg x (12 ms$^{-1}$)$^2$ = 72 J

Second part: KE = 0.5 x mass x velocity$^2$ = 0.5 x 2 kg x (8 ms$^{-1}$)$^2$ = 64 J

The total kinetic energy of these two parts is:

Total KE = 72 J + 64 J = 136 J

According to the law of conservation of energy, the kinetic energy of the third part must be equal to the difference between the total kinetic energy of the first two parts and the initial kinetic energy of the rock. The initial kinetic energy of the rock was zero, so we have:

KE of third part = Total KE - 0 = 136 J

Using the formula for kinetic energy, we can write:

0.5mv$^2$ = 136 J

Solving for "v", we get:

v = $\sqrt{\frac{272}{m}}$

Substituting

Perpendicular to the original direction of B
d)Cannot be determined

Answer:
c) Perpendicular to the original direction of B

Explanation:
We can solve this problem using conservation of momentum and conservation of kinetic energy.

Before the collision, the total momentum of the system is:

p = m2v

Since sphere A is at rest, its momentum is zero.

The total kinetic energy of the system before the collision is:

K = (1/2)m2v^2

After the collision, the spheres move in different directions. Let the velocity of sphere A be u and the velocity of sphere B be w. Then, the total momentum of the system after the collision is:

p' = m1u + m2w

Since sphere B moves perpendicular to its original direction, we can write:

w = 2v

Using conservation of momentum, we have:

m2v = m1u + m2(2v)

Simplifying, we get:

u = (m2 - 2m1)v / m1

Now, using conservation of kinetic energy, we have:

(1/2)m1u^2 + (1/2)m2w^2 = (1/2)m2v^2

Substituting the values of u and w, we get:

(m2 - 2m1)v^2 = m1u^2

Simplifying, we get:

u = sqrt((m2 - 2m1)/m1) v

Since m1 and m2 are positive, (m2 - 2m1)/m1 is negative. Therefore, u is imaginary, which means that sphere A moves in a direction perpendicular to the original direction of sphere B. Hence, the answer is option c) Perpendicular to the original direction of B.

Gravitational and electrical forces are conservative. Friction is non-conservative because the amount of work done by friction depends on the path. One can associate a potential energy with a conservative force but not with a non-conservative force.

Given:
Velocity of water, v = 2 m/s
Mass per unit length of water, m = 100 kg/m

To find:
Power of the engine

Formula used:
Power = Force × Velocity

Force = mass × acceleration = mass × change in velocity / time

Change in velocity = final velocity - initial velocity = v - 0 = v

Time taken to move out of the pipe, t = length of the pipe / velocity of water

Let the length of the pipe be L.

Therefore, t = L / v

Force = m × v / t = m × v² / L

Power = Force × Velocity = m × v² / L × v = m × v² / L

Substituting the given values, we get:

Power = 100 kg/m × (2 m/s)² / L = 400 W

Therefore, the power of the engine is 400 W.

1. Momentum: momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
2. The unit of momentum (P) is kg m/s.
3. Dimension: [MLT-1]
4. Law of conservation of Momentum: A conservation law stating that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.
5. P₁ = P₂
6. m₁ v₁ = m₂ v₂
7. Where, P₁ = initial momentum of system, P₂ = final momentum of system, m₁ = mass of first object, v₁ = velocity of first object, m₂ = mass of second object and v₂ = velocity of second object.

-1. The radius of the cylinder is 0.5 m. Calculate the kinetic energy of the cylinder.

First, we need to calculate the angular velocity of the cylinder, which can be found using the formula:

v = ωr

where v is the linear velocity, ω is the angular velocity, and r is the radius of the cylinder. Rearranging this formula, we get:

ω = v/r

Substituting the given values, we get:

ω = 4/0.5 = 8 rad/s

The kinetic energy of the cylinder can be calculated using the formula:

KE = (1/2)Iω^2 + (1/2)mv^2

where I is the moment of inertia of the cylinder, m is the mass of the cylinder, and v is the linear velocity.

The moment of inertia of a solid cylinder is given by:

I = (1/2)mr^2

Substituting the given values, we get:

I = (1/2)×3×(0.5)^2 = 0.375 kg m^2

Substituting all the values in the formula for KE, we get:

KE = (1/2)×0.375×8^2 + (1/2)×3×4^2
= 12 + 24
= 36 J

Therefore, the kinetic energy of the cylinder is 36 Joules.

Given Data
- Mass of engine (m) = 100,000 kg
- Gradient (θ) = 5°
- Speed (v) = 100 m/hr = 100/3600 m/s = 0.02778 m/s
- Coefficient of friction (μ) = 0.1
- Efficiency (η) = 4% = 0.04
- Heat energy from 1 kg of coal = 50,000 J
Calculating Forces
- Weight of the engine (W):
W = m * g = 100,000 kg * 9.81 m/s² = 981,000 N
- Component of weight along the slope (W_slope):
W_slope = W * sin(θ) = 981,000 N * sin(5°) ≈ 85,757 N
- Normal force (N):
N = W * cos(θ) ≈ 981,000 N * cos(5°) ≈ 979,365 N
- Frictional force (F_friction):
F_friction = μ * N = 0.1 * 979,365 N ≈ 97,937 N
Total Force Required (F_total)
F_total = W_slope + F_friction ≈ 85,757 N + 97,937 N ≈ 183,694 N
Power Required
- Power (P):
P = F_total * v = 183,694 N * 0.02778 m/s ≈ 5,100 W
Energy Required in One Hour
- Energy (E) in one hour (3600 seconds):
E = P * t = 5,100 W * 3600 s ≈ 18,396,000 J
Coal Required
- Energy output from coal:
Energy from burning coal = mass * 50,000 J/kg
- Using efficiency:
Effective energy from coal = η * Energy from coal
Setting these equal:
18,396,000 J = 0.04 * (mass * 50,000 J)
Solving for mass:
mass = 18,396,000 J / (0.04 * 50,000 J) ≈ 9154 kg
Conclusion
The engine needs to burn approximately 9154 kg of coal in one hour to maintain its operation on the slope. Thus, the correct answer is option D.

Understanding the Problem
The body is being lowered onto a platform that is moving horizontally at a speed of 4 m/s. The coefficient of friction between the body and the platform is 0.2. We need to determine how far the body will slide relative to the platform.
Calculating the Forces
- The weight of the body (W) can be calculated using the equation:
W = m * g
Where g = 10 m/s².
- The frictional force (F_f) acting on the body is given by:
F_f = μ * N
Where μ is the coefficient of friction (0.2), and N is the normal force, which equals the weight of the body (W).
Setting Up the Equations
- The maximum frictional force before sliding occurs:
F_f = 0.2 * W = 0.2 * m * g
- The body will slide when the friction is overcome by the inertial force due to the platform's movement.
Determining the Distance
- The critical speed at which the body can remain at rest on the platform is determined by the maximum frictional force. If the platform moves at 4 m/s, we can calculate the distance (d) it will slide before coming to rest relative to the platform.
- Applying the equation of motion under constant acceleration for sliding:
d = (v^2)/(2 * a)
where v = initial relative speed (4 m/s) and a = deceleration due to friction.
- Deceleration (a) due to friction is:
a = F_f/m = μ * g = 0.2 * 10 = 2 m/s².
- Substituting the values:
d = (4^2)/(2 * 2) = 16/4 = 4 m.
Conclusion
The body will slide a distance of 4 meters relative to the platform before stopping, confirming that the correct answer is option 'C'.