Work & Energy MCQs set 2 for UHS Punjab MDCAT Physics — 20 solved questions.
Q1. A 50 N force is applied to a box, causing it to move 8 m along a frictionless surface. What is the work done?
Answer: 400 J
Explanation: Work done = Force × Distance = 50 N × 8 m = 400 J. Option C fails because it's not the correct product.
Q2. A 2 kg block is lifted 5 m vertically. What is the gain in gravitational potential energy? (g = 10 m/s²)
Answer: 100 J
Explanation: Gain in GPE = mgh = 2 kg × 10 m/s² × 5 m = 100 J. Option A fails due to incorrect calculation.
Q3. A car accelerates from rest to 20 m/s in 4 s. What is the gain in kinetic energy if its mass is 1500 kg?
Answer: 300 kJ
Explanation: Gain in KE = ½mv² = ½ × 1500 kg × (20 m/s)² = 300 kJ. Option B fails due to miscalculation.
Q4. A force of 20 N acts on a body at an angle of 60° to its direction of motion. What is the work done in moving the body 10 m?
Answer: 100 J
Explanation: Work done = F × d × cos(θ) = 20 N × 10 m × cos(60°) = 100 J. Option A fails due to incorrect angle application.
Q5. A 1000 kg elevator is lifted 20 m. What is the work done against gravity? (g = 10 m/s²)
Answer: 200 kJ
Explanation: Work done = mgh = 1000 kg × 10 m/s² × 20 m = 200 kJ. Option A fails due to miscalculation.
Q6. A spring with a spring constant of 500 N/m is compressed by 0.2 m. What is the elastic potential energy stored?
Answer: 10 J
Explanation: Elastic PE = ½kx² = ½ × 500 N/m × (0.2 m)² = 10 J. Option A fails due to incorrect formula application.
Q7. A 5 kg object is moving at 4 m/s. What is its kinetic energy?
Answer: 40 J
Explanation: KE = ½mv² = ½ × 5 kg × (4 m/s)² = 40 J. Option B fails due to incorrect calculation.
Q8. A machine lifts a 200 N weight to a height of 10 m in 5 s. What is the power output of the machine?
Answer: 400 W
Explanation: Power = Work / Time = (200 N × 10 m) / 5 s = 400 W. Option B fails due to miscalculation.
Q9. A body of mass 10 kg is moving with a velocity of 5 m/s. What is the work done to stop it?
Answer: -125 J
Explanation: Work done = ΔKE = 0 - ½ × 10 kg × (5 m/s)² = -125 J. Option C fails because it's positive.
Q10. A force of 100 N is applied to a stationary block of mass 20 kg. If the block doesn't move, what is the work done?
Answer: 0 J
Explanation: Work done = 0 since there's no displacement. Option B fails because it assumes displacement.
Q11. A 2 kg object is dropped from a height of 10 m. What is its kinetic energy just before hitting the ground? (g = 10 m/s²)
Answer: 200 J
Explanation: KE = mgh = 2 kg × 10 m/s² × 10 m = 200 J. Option A fails due to miscalculation.
Q12. A car of mass 1200 kg accelerates from 10 m/s to 20 m/s. What is the increase in its kinetic energy?
Answer: 180 kJ
Explanation: ΔKE = ½m(v₂² - v₁²) = ½ × 1200 kg × ((20 m/s)² - (10 m/s)²) = 180 kJ. Option B fails due to incorrect calculation.
Q13. A 50 g bullet is fired at 400 m/s. What is its kinetic energy?
Answer: 4000 J
Explanation: KE = ½mv² = ½ × 0.05 kg × (400 m/s)² = 4000 J. Option B fails due to miscalculation.
Q14. A 10 kg block is pulled 5 m up a frictionless incline with a force of 50 N parallel to the incline. What is the work done by this force?
Answer: 250 J
Explanation: Work done = F × d = 50 N × 5 m = 250 J. Option A fails due to incorrect calculation.
Q15. A 500 N force is applied to a car, causing it to move 20 m. If the force is applied at an angle of 30° to the direction of motion, what is the work done?
Answer: 8660 J
Explanation: Work done = F × d × cos(θ) = 500 N × 20 m × cos(30°) = 8660 J. Option A fails due to incorrect angle application.
Q16. A 20 kg child on a swing is pulled back to a height of 2 m above the lowest point. What is the child's gravitational potential energy at this height? (g = 10 m/s²)
Answer: 400 J
Explanation: GPE = mgh = 20 kg × 10 m/s² × 2 m = 400 J. Option B fails due to miscalculation.
Q17. A machine has an efficiency of 80% and does 400 J of work. What is the input energy to the machine?
Answer: 500 J
Explanation: Efficiency = Output / Input; 0.8 = 400 J / Input; Input = 400 J / 0.8 = 500 J. Option A fails due to incorrect efficiency application.
Q18. A 100 g ball is thrown upwards with an initial velocity of 20 m/s. What is its kinetic energy at the highest point?
Answer: 0 J
Explanation: At the highest point, velocity = 0, so KE = 0. Option B fails because it assumes some velocity remains.
Q19. A force of 200 N is applied to a 50 kg block on a frictionless surface, causing it to accelerate. What is the work done by this force in 4 s if the block starts from rest?
Answer: 3200 J
Explanation: a = F/m = 200 N / 50 kg = 4 m/s²; d = ½at² = ½ × 4 m/s² × (4 s)² = 32 m; Work = F × d = 200 N × 32 m = 6400 J, but using the work-energy theorem: Work = ΔKE = ½ × 50 kg × (16 m/s)² = 3200 J (since v = u + at = 0 + 4 × 4 = 16 m/s). Option A fails due to incorrect time application.
Q20. A 2 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²)
Answer: 10 m/s
Explanation: mgh = ½mv²; v = √(2gh) = √(2 × 10 m/s² × 5 m) = 10 m/s. Option B fails due to miscalculation.