Railway Science (Physics-Work, Energy and Power) (Part-V)

Total Questions: 16

1. A body of 4.0 kg is lying at rest. Under the action of a constant force, it gains a speed of 5 m/s. The work done by the force will be_________ . [RRB ALP Tier - I (09/08/2018) Afternoon]

Correct Answer: (b) 50 J
Solution:

Work energy theorem - Sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body.Example - Work done by all the forces =

2. What will be the value of the kinetic energy (Eₖ) of a moving body with mass m, if its speed is doubled from v to 2v ? [RRB ALP Tier - I (09/08/2018) Evening]

Correct Answer: (b) 4Eₖ
Solution:

3. A ball of 0.1 kg is dropped from rest. When it falls through a distance of 2 m, the work done by the force of gravity is (g = 9.8m/𝑠²): [RRB ALP Tier - I (10/08/2018) Evening]

Correct Answer: (d) 1.96 J
Solution:

Work done = F × S, Where F = force acting on the object and S =displacement.
Now, Force (F) = ma, ⇒ F = 0.1 kg × 9.8 m/s².
Thus, Work = F × S = 0.1 × 9.8 × 2 = 1.96 J.

4. A ball is dropped from a height of 10 m. It strikes the ground and rebounds up to a height of 2.5 m. During the collision, the percent loss in the kinetic energy is: [RRB ALP Tier - I (13/08/2018) Morning]

Correct Answer: (c) 75%
Solution:

According to the law of conservation of mechanical energy: Initial kinetic energy = Potential Energy = mgh = 10
Final kinetic energy = mgh₁ = 2.5
Change in kinetic energy = mg (h - h₁) = 10 - 2.5 = 7.5
Percentage change in kinetic energy =
𝑚𝑔 (ℎ−ℎ1)/𝑚𝑔ℎ × 100 = 7.5/10 × 100 = 75%.

5. If the kinetic energy of a body becomes 256 times its initial value, then the new linear momentum will be: [RRB ALP Tier - I (14/08/2018) Morning]

Correct Answer: (c) 16 times the initial value
Solution:

6. An object weighing 20 kg is raised through a height of 2 m. What will be the work done by the force of gravity in this process? (Take g = 10 m/s²) [RRB ALP Tier - I (14/08/2018) Afternoon]

Correct Answer: (c) 400 J
Solution:

Work done by the force of gravity
= Gravitational Potential Energy = m × g × h.
Gravitational Potential Energy = m (mass) × g (acceleration) × h (height) 2 × 10 × 20 = 400 J.
Hence, in this process work done by the force of gravity is 400 J.

7. Mohan, having a mass of 40 kg, runs up a staircase of 50 steps in 10 s. If the height of each step is 15 cm, then what is his power? (Take g = 10 m/s²) [RRB ALP Tier - I (14/08/2018) Evening]

Correct Answer: (b) 300 W
Solution:

Power: The rate of  work done is called power. Power (P) =
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 (𝑊)/𝑇𝑖𝑚𝑒 (𝑡). Potential energy: The energy of a body due to its position. Potential energy (PE) = m g h.
Height = 15cm, time (t) = 10s, Mass (m) = 40 kg, g = 10 m/s.
Total Height of staircase (h) = 50 × 15 = 750 cm = 7.5 m
Potential energy (PE) = m g h = 40 × 10 × 7.5 = 3000 J.
Work done (W) = PE = 3000 J.
Power (P) = 𝑊/𝑡 = 3000/10 = 300 Watt.

8. What is the amount of work done when a body moves under a force of 1N at a distance of 1 metre in the direction of the force? [RRB ALP Tier - I (20/08/2018) Afternoon]

Correct Answer: (c) 1 Joule
Solution:

Joule is the SI unit of work. Work is a scalar quantity. Force is a vector quantity, SI unit newton (N).
Given : Force (F) = 1 N, Distance (d) = 1 m
⇒ work done (W) =
F × d = 1 N × 1 m = 1 Joule
Thus, when a force of 1 N moves a body through a distance of 1 m in its own direction then the work done is 1 Joule.

9. An object having a mass 'm' moving with a velocity ‘v' possesses a kinetic energy ‘K’. If its velocity is doubled, its kinetic energy will become: [RRB ALP Tier - I (20/08/2018) Evening]

Correct Answer: (b) 4 K
Solution:

10. Work done by an object on application of a force would be zero if the displacement of the object is: [RRB ALP Tier - I (21/08/2018) Evening]

Correct Answer: (b) zero
Solution:

W = F × S, Where w = work done, F = force, S = displacement
W = F × S. Let F = 1, W= 0,
0 = 1 × S ⇒ ∴ S = 0