Work, Energy and Power
NIOS · Class 12 · Physics
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A force of 10 N is applied on a block at an angle of 60° with the horizontal. If the block moves 4 m along the horizontal surface, what is the work done by the force?
A body of mass 5 kg is moving with a velocity of 6 m/s. A net force acts on it and its velocity increases to 10 m/s. What is the work done by the net force?
A pump lifts 200 kg of water to a height of 10 m every minute. What is the power of the pump? (g = 10 m/s²)
A spring with force constant k = 200 N/m is compressed by 0.1 m. What is the elastic potential energy stored in the spring?
Sample Questions
A block is moved from point A to point B along two different paths on a rough surface. Which of the following statements is correct?
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Work done by gravity is the same along both paths.
Step 1: Gravity is a conservative force; its work depends only on the vertical displacement (height difference), not on the path. Step 2: Friction is a non-conservative force; its work depends on the length and nature of the path. Step 3: Since friction does negative work and is path-dependent, total mechanical energy is NOT conserved. Option A is wrong because friction is path-dependent. Option C is wrong because friction dissipates energy. Option D is the definition of a conservative force, which friction is NOT.
Two identical balls A and B have masses m each. A moves with velocity v and B is at rest. They undergo a perfectly elastic head-on collision. What are the velocities of A and B after the collision?
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A: 0, B: v
Step 1: In an elastic collision between identical masses, velocities are exchanged. Step 2: Using the result for equal masses (mA = mB = m) with vBi = 0: vAf = vBi = 0 and vBf = vAi = v. Step 3: A comes to rest and B moves with A's initial velocity v. This is verified by checking both momentum (mv = mv ✓) and kinetic energy (½mv² = ½mv² ✓) conservation. Option A is the result of an inelastic collision. Option C means no collision happened. Option D violates energy conservation.
A ball of mass 2 kg is dropped from a height of 20 m. What is its kinetic energy just before it hits the ground? (g = 10 m/s²)
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400 J
Step 1: By conservation of energy, all PE at the top converts to KE at the bottom. Step 2: PE at top = mgh = 2 × 10 × 20 = 400 J. Step 3: KE just before hitting ground = 400 J (since PE = 0 at ground level). Option A (200 J) uses g = 5 or h = 10 incorrectly. Option B (20 J) only uses mh. Option D (800 J) doubles the result.
A person carries a 10 kg bag horizontally at a constant velocity for 5 m on a level floor. How much work is done by the person against gravity?
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0 J
Step 1: Work done = Fd cosθ, where θ is the angle between force and displacement. Step 2: Gravity acts vertically downward; displacement is horizontal. Step 3: θ = 90°, so cos 90° = 0. Step 4: W = mgh cos 90° = 0 J. The person does no work against gravity because there is no vertical displacement. Options A and B incorrectly assume vertical displacement. Option D is half of mgs which has no physical meaning here.
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