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

Total Questions: 50

41. A porter lifts a luggage of 12 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage (g = 10 ms⁻²). [RRB Group D 04/10/2018 (Morning)]

Correct Answer: (c) 180 J
Solution:

∵ Work done = change in potential energy
= mgΔh, where m is the mass lifted, g is the acceleration due to gravity, h is the change in height.
∴ W = mgΔh = 12 × 10 × 1.5 = 180 J.

42. A force of 10 N is acting on an object. The object is displaced 5 m in the direction of the force. So the work done is: [RRB Group D 04/10/2018 (Morning)]

Correct Answer: (c) 50 J
Solution:

Given that, Force (F) = 10 N, Displacement (s) = 5 m, Work done = ?
∴ Work done = Force × Displacement = 10 × 5 = 50 J

43. An object of mass 12 kg is kept at a certain height from the ground. If the potential energy of the object is 600 J, then find the height of the object with respect to the ground. Given g = 10 ms⁻² [RRB Group D 04/10/2018 (Morning)]

Correct Answer: (d) 5 m
Solution:

Given that, mass (m) = 12 kg, Potential energy (P.E) = 600 J, Acceleration due to gravity (g) = 10 ms⁻², Height (h) = ?
We know that, Potential energy (P.E)
= m × g × h
⇒ 600 = 12 × 10 × h ⇒ h = 5 m.

44. A porter lifts a luggage of 14 kg from the ground, and puts it on his head, 1.5m above the ground. Calculate the work done by him on the luggage. [RRB Group D 04/10/2018 (Afternoon)]

Correct Answer: (a) 210 J
Solution:

Given that : Mass of luggage (m) = 14 kg,
Height of luggage (h) = 1.5 m
Work done = m × g × h = 14 × 10 × 1.5
[since, acceleration due to gravity(g) = 10 ms⁻²] = 210 J.

45. Find the kinetic energy of a ball of mass 2 kg moving with a speed of 30 ms⁻¹: [RRB Group D 04/10/2018 (Afternoon)]

Correct Answer: (b) 900 J
Solution:

46. Calculate the potential energy acquired by a 15 kg hammer when it is raised to a height of 10 m (Given g = 10 ms⁻²). [RRB Group D 05/10/2018 (Morning)]

Correct Answer: (a) 1500 J
Solution:

Given, mass= 15 kg, g=10 m/s² , height = 10 m.
Potential energy (PE) = Mass × acceleration due to gravity × height of the object.
⇒ PE = 15 × 10 × 10 = 1500 J.

47. A porter lifts a luggage of 13 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage. (g = 10 ms⁻²) [RRB Group D 05/10/2018 (Morning)]

Correct Answer: (d) 195 J
Solution:

Work Done = mass (m) × acceleration due to gravity (g) × height (h) = 13 × 10 × 1.5 m = 195 J.

48. A worker takes a 15 kg piece from the ground and puts it on his head, 1.0 m above the ground. Calculate the work done by him on the goods. (g = 10m𝑠⁻² ) [RRB Group D 05/10/2018 (Afternoon)]

Correct Answer: (b) 150 J
Solution:

Given that, Mass (m) = 15 kg, Gravity (g) = 10 ms⁻², Height (h) = 1 m.
We know that, Work= m × g × h
⇒ Work= 15 × × 10 × 1 = 150 J.

49. A force of 20 N is acting on an object. The object is displaced through 4 meters in the direction of the force. Then the work done is: [RRB Group D 05/10/2018 (Afternoon)]

Correct Answer: (d) 80 J
Solution:

Given that, Force = 20 N, Displacement = 4 m.
We know that, Work done = force × displacement,
⇒ Work done= 20 × 4 = 80 J.

50. A porter lifts a luggage of 10 kg from the ground and puts it on his head 1.1 m above the ground. Calculate the work done by him on the luggage. (g = 10 ms⁻²) [RRB Group D 05/10/2018 (Evening)]

Correct Answer: (d) 110 J
Solution:

Given, Mass of the luggage = 10 kg,
Height = 1.1 m.
Potential Energy = m × g × h
= 10 × 10 × 1.1= 110 J.
The work done by the porter is 110 J.