A lift of mass M=500 kg is descending with speed of 2 ms⁻¹. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 ms⁻². The kinetic energy of the lift at the end of fall through to a distance of 6 m will be _____________ kJ.

Solution

Physics sir · 4 steps

Do not use g here: the net force is fixed by the given acceleration, so net work is just Ma × s.

1. Starting energy. The lift is already moving at 2 m s⁻¹, so it starts with Kᵢ = 1/2(500)(2)² = 1000 J.
2. Net work done. Gravity and the slipping cable together give a net acceleration of 2 m s⁻². So the net force is F = Ma = 500 × 2 = 1000 N, and over 6 m it does W = 1000 × 6 = 6000 J of net work.

   K_f = Kᵢ + Wₙₑₜ = 1000 + 6000 = 7000 J

3. Quick check. With v² = u² + 2as = 4 + 24 = 28 m² s⁻², you get K_f = 1/2(500)(28) = 7000 J, the same result.
4. So. The kinetic energy after falling 6 m is 7000 J, which is 7 kJ.

The common slip: 6. Forgetting the initial speed of 2 m s⁻¹ counts only the work done, 6000 J, which gives 6 kJ instead of 7.

The idea to keep. The work-energy theorem uses the net force. When the acceleration is given, net force is Ma, and the starting kinetic energy must be added to the work done.

Answer from the NTA final answer key. Solution by PrepWiser.