NEET Physics: Work, Energy and Power questions with solutions

Q1. Assertion A load when lifted vertically up requires greater effort than to roll the same load up the ramp. Reason The work done in lifting the load vertically up is less.

1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
3. Assertion is correct but Reason is incorrect
4. Both Assertion and Reason are incorrect

Answer: Assertion is correct but Reason is incorrect

Lifting vertically needs a larger force than pushing the same load up a ramp, so the assertion is correct. But the reason is wrong because the work done in lifting vertically is not less; ideally, it is the same as the work done along the ramp (ignoring friction).

Q2. The graph between the resistive force acting on a body and the position of the body travelling in a straight line is shown in the figure. The mass of the body is \( 25 \mathrm{kg} \) and initial velocity is 2 m/s.When the distance covered by the bodyi \( s 4 \mathrm{m}, \) its kinetic energy would be

1. 50
2. 40 J
3. 20
4. 10

Answer: 40 J

The resistive force does negative work, and that work equals the area under the force–position graph over 0 to 4 m. Subtracting this loss from the initial kinetic energy gives the final kinetic energy, which comes out to 40 J.

Q3. Potential energy increases with the increase in :

1. work
2. Force
3. speed
4. Position

Answer: Position

Potential energy is stored energy due to an object's position in a force field, such as gravity. As position changes to a higher or more favorable location, potential energy increases.

Q4. Three different objects of masses m1, m2, and m3 are allowed to fall from rest and from the same point O along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of

1. (a) m1 : m2 : m3
2. (b) m1 : 2m2 : 3m3
3. (c) 1 : 1 : 1
4. (d) 1/m1 : 1/m2 : 1/m3

Answer: (c) 1 : 1 : 1

The speed of an object falling under gravity depends only on the height and not on its mass, as per the equation v = √(2gh). Since all objects fall from the same height, their speeds will be the same, giving a ratio of 1:1:1.

Q5. What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. √(gR)
2. √(2gR)
3. √(3gR)
4. √(5gR)

Answer: √(5gR)

To complete the vertical loop, the body must have sufficient velocity at the top of the loop to maintain contact with the track. At the top, the centripetal force is provided by the gravitational force and the normal force. For minimum velocity, the normal force becomes zero, and the centripetal force equals the gravitational force. Using energy conservation and centripetal force conditions, the minimum velocity at the bottom is derived as √(5gR).

Q6. A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then:

1. E₁ = E₂
2. E₁ < E₂
3. E₁ = m₁/m₂
4. E₁ > E₂

Answer: E₁ < E₂

Since both particles have the same momentum (p = m*v), the velocity of the lighter particle (m₂) will be greater than that of the heavier particle (m₁). Kinetic energy is proportional to the square of velocity (KE = p²/2m). Therefore, the lighter particle (m₂) will have a higher kinetic energy (E₂ > E₁).

Q7. A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of:

1. 1 : √2
2. √2 : 1
3. 1 : 4
4. 1 : 2

Answer: 1 : 2

The kinetic energy of an object is given by KE = (1/2)mv². Since both balls fall the same distance, they will have the same velocity due to the same acceleration (gravity). Thus, the ratio of their kinetic energies depends only on their masses, which is 2:4 or 1:2.

Q8. When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be:

1. 25 U
2. U/5
3. 5 U
4. 10 U

Answer: 25 U

The potential energy stored in a spring is proportional to the square of the extension (U ∝ x²). If the extension increases from 2 cm to 10 cm, the ratio of extensions is 10/2 = 5. Therefore, the potential energy increases by a factor of 5² = 25, making the new potential energy 25U.

Q9. Two bodies with kinetic energies in the ratio 4 : 1 are moving with equal linear momentum. The ratio of their masses is

1. 1 : 2
2. 1 : 1
3. 4 : 1
4. 1 : 4

Answer: 1 : 2

Kinetic energy is given by KE = p²/2m, where p is momentum and m is mass. Since the momenta are equal, the ratio of kinetic energies (KE₁/KE₂ = 4/1) implies m₂/m₁ = 2/1. Thus, the ratio of their masses is 1:2.

Q10. A rubber ball is dropped from a height of 5m on a plane, where the acceleration due to gravity is not shown. On bouncing it rises to 1.8m. The ball loses its velocity on bouncing by a factor of

1. 16/25
2. 2/5
3. 3/5
4. 9/25

Answer: 3/5

The velocity of the ball just before hitting the ground is proportional to the square root of the height it falls from, and the velocity after bouncing is proportional to the square root of the height it rises to. The ratio of velocities is √(1.8/5) = 3/5, so the ball loses its velocity by a factor of 3/5.

Q11. Two bodies of masses m and 4m are moving with equal K.E. The ratio of their linear momenta is

1. 4 : 1
2. 1 : 1
3. 1 : 2
4. 1 : 4

Answer: 1 : 2

Kinetic energy is given by K.E. = p²/2m, where p is momentum. For equal K.E., p ∝ √m. Thus, the ratio of their momenta is √m : √(4m) = 1 : 2.

Q12. The kinetic energy acquired by a mass (m) in travelling distance (s) starting from rest under the action of a constant force is directly proportional to

1. 1/√m
2. 1/m
3. √m
4. m⁰

Answer: m⁰

The work-energy theorem states that the work done by the force is equal to the change in kinetic energy. Since the force is constant, work done is proportional to the distance (s), and kinetic energy is independent of the mass (m). Thus, it is proportional to m⁰, which equals 1.

Q13. If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is

1. 50%
2. 100%
3. 125%
4. 200%

Answer: 125%

Kinetic energy (K) is proportional to the square of momentum (p), i.e., K ∝ p². If momentum increases by 50%, the new momentum becomes 1.5p. The new kinetic energy becomes (1.5p)² = 2.25p², which is a 125% increase from the original kinetic energy.

Q14. Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:

1. 1/9
2. 8/9
3. 4/9
4. 5/9

Answer: 8/9

In a head-on elastic collision, the fraction of energy lost by body A can be calculated using the formula for energy transfer in elastic collisions. For masses 4m and 2m, the fraction of energy lost by A is 8/9.

Q15. If E is the kinetic energy of a body of mass m moving with velocity v, then:

1. E = mv²
2. E = p² / 2m
3. E = p² / m
4. E = 2mE

Answer: E = p² / 2m

The kinetic energy of a body is given by E = (1/2)mv². Momentum p is given by p = mv. Substituting p = mv into the kinetic energy formula, we get E = p² / 2m.

Q16. Since height is the same for both balls, their velocities on reaching the ground will be the same. If K.E.₁ = 1/2 m₁v₀² and K.E.₂ = 1/2 m₂v₀², then:

1. K.E.₁ = K.E.₂
2. K.E.₁ > K.E.₂
3. K.E.₁ < K.E.₂
4. K.E.₁ = 1/2 K.E.₂

Answer: K.E.₁ > K.E.₂

The kinetic energy depends on both mass and velocity. Since the velocities are the same for both balls, the ball with the greater mass (m₁ or m₂) will have greater kinetic energy. Thus, if m₁ > m₂, K.E.₁ > K.E.₂.

Q17. If k be the spring constant, then:

1. U = 1/2 k × x² = 2k
2. Ufinal = 1/2 k × x(10)² = 50k
3. New K.E., E' = E' + 3E = 4E
4. p' / p = √(2m × 4E / 2mE') = 2

Answer: p' / p = √(2m × 4E / 2mE') = 2

Option D is correct because the ratio of momentum (p'/p) is derived using the relationship between kinetic energy and momentum, where p = √(2mE). The ratio simplifies to 2 as given.

Q18. According to principle of conservation of energy, Loss in potential energy = Gain in kinetic energy. If h1 and h2 are initial and final heights, then v1 = √2gh1 and v2 = √2gh2.

1. (a) mgh = 1/2 mv^2 => v = √2gh
2. (b) If h1 and h2 are initial and final heights, then v1 = √2gh1 and v2 = √2gh2
3. (c) Loss in velocity, Δv = v1 - v2 = √2gh1 - √2gh2
4. (d) Fractional loss in velocity, Δv / v1 = √2gh1 - √2gh2 / √2gh1

Answer: (b) If h1 and h2 are initial and final heights, then v1 = √2gh1 and v2 = √2gh2

Option (b) correctly states the relationship between velocity and height derived from the conservation of energy principle, where the potential energy converts into kinetic energy.

Q19. According to principle of conservation of energy, Loss in potential energy = Gain in kinetic energy. If h1 and h2 are initial and final heights, then v1 = √2gh1 and v2 = √2gh2.

1. (a) m1v1^2 = m2v2^2
2. (b) m1v1^2 / m2v2^2 = p1^2 / p2^2
3. (c) m2 / m1 = 4 x p2^2 / p1^2
4. (d) m1 / m2 = 1 / 4

Answer: (c) m2 / m1 = 4 x p2^2 / p1^2

Using the principle of conservation of energy, the relationship between mass, velocity, and momentum can be derived. Since momentum p = mv, the ratio of masses and momenta leads to the given expression m2/m1 = 4 × (p2^2 / p1^2).

Q20. K.E. = 1/2 mv^2 = Work done = F x d = constant.

1. (a) Further, v^2 = u^2 + 2as = 0 + 2ad = 2(F/m)d
2. (b) K.E. = 1/2 mv^2 = Work done = F x d = constant
3. (c) K.E. acquired = Work done = F x d = constant
4. (d) All of the above

Answer: (d) All of the above

All the options are correct as they represent valid relationships derived from the work-energy theorem and kinematic equations. The kinetic energy acquired by an object is equal to the work done on it, and the equations are consistent with the principles of mechanics.