Work, Power and Energy Quiz - ICSE Class 11 Physics
Free practice quiz: Work, Power and Energy (ICSE Class 11 Physics). 38 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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A particle of mass m is projected vertically upwards with an initial velocity v. What is the work done by gravity when the particle reaches its maximum height?
- 0
- mv²/2
- -mv²/2
- mg(v²/2g)
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Answer: C. -mv²/2
The initial kinetic energy is KE_i = mv²/2. At the maximum height, the velocity is zero, so the final kinetic energy is KE_f = 0. By the work-energy theorem, the work done by gravity (W_g) is equal to the change in kinetic energy: W_g = KE_f - KE_i = 0 - mv²/2 = -mv²/2. Alternatively, the maximum height reached is h = v²/2g. The work done by gravity is W_g = -mgh = -mg(v²/2g) = -mv²/2.
A 5 kg object is lifted vertically upwards by a force of 100 N. If the object starts from rest and reaches a height of 10 m, what is the work done by the applied force?
- 500 J
- 1000 J
- 490 J
- 50 J
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Answer: B. 1000 J
The work done by a constant force is given by W = Fd, where F is the magnitude of the force and d is the displacement in the direction of the force. Here, the applied force is F = 100 N and the vertical displacement is d = 10 m. Therefore, the work done by the applied force is W = 100 N * 10 m = 1000 J. Note that the weight of the object is mg = 5 kg * 9.8 m/s² = 49 N. The net force is 100 N - 49 N = 51 N. The work done against gravity is 490 J. The work done by the applied force is 1000 J.
A 2 kg mass is moving with a velocity of 3 m/s. What is its kinetic energy?
- 3 J
- 6 J
- 9 J
- 18 J
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Answer: C. 9 J
Kinetic energy (KE) is given by the formula KE = ½mv², where m is the mass and v is the velocity. Here, m = 2 kg and v = 3 m/s. So, KE = ½ * 2 kg * (3 m/s)² = 1 kg * 9 m²/s² = 9 J.
A block of mass 4 kg is pulled by a force of 16 N on a smooth horizontal surface. If it moves a distance of 5 m, what is the work done by the force?
- 16 J
- 64 J
- 80 J
- 320 J
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Answer: C. 80 J
Work done by a constant force is given by W = F * d. Here, F = 16 N and d = 5 m. Since the force is applied horizontally and the displacement is horizontal, the angle between them is 0°, and cos(0°) = 1. Therefore, W = 16 N * 5 m = 80 J. The mass of the block is not required for this calculation.
A spring is compressed by 0.2 m from its equilibrium position. If the work done to compress it is 8 J, what is the spring constant?
- 100 N/m
- 200 N/m
- 400 N/m
- 800 N/m
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Answer: C. 400 N/m
The work done in compressing a spring from its equilibrium position is given by W = ½kx², where k is the spring constant and x is the compression. We are given W = 8 J and x = 0.2 m. So, 8 J = ½ * k * (0.2 m)². 8 = ½ * k * 0.04. 8 = k * 0.02. k = 8 / 0.02 = 400 N/m.