A lift (elevator) cabin and its passengers have a total mass of 850 kg. Starting from rest, the lift accelerates uniformly upward. In a time of 5.0 s it rises through a vertical height of 6.0 m and reaches a final speed of 2.5 m s^-1. Assume no energy is lost to resistive forces…

Answer: ((a)) gain in gravitational PE = 50031 J≈5.00×10^4 J (from 850×9.81×6.0) ((b)) gain in kinetic energy = 2656.25 J≈2.66×10^3 J (from 1/2×850×2.5^2) ((c)) total work done by the motor = 52687.25 J≈5.27×10^4 J (from 50031+2656.25) ((d)) average power output ≈1.05×10^4 W (from 52687.25/5.0). Part (a): Gain in gravitational potential energy The gain in gravitational PE depends on the mass and the vertical height risen: Ep = mg h = 850×9.81×6.0 Working in two steps: 850×9.81=8338.5, then 8338.5×6.0=50031 J. Check by recomputing…

A Level Physics 9702: Energy and Average Power of an Accelerating Lift

A lift (elevator) cabin and its passengers have a total mass of $850\text{ kg}$ . Starting from rest, the lift accelerates uniformly upward. In a time of $5.0\text{ s}$ it rises through a vertical height of $6.0\text{ m}$ and reaches a final speed of $2.5\text{ m s}^{-1}$ . Assume no energy is lost to resistive forces such as friction in the lift mechanism.

Take $g = 9.81\text{ m s}^{-2}$ .

(a) Calculate the gain in gravitational potential energy of the lift during this time. [2]

(b) Calculate the gain in kinetic energy of the lift during this time. [2]

(d) Calculate the average power output of the lift motor during this time. [2]

Part (a): Gain in gravitational potential energy

The gain in gravitational PE depends on the mass and the vertical height risen: $\Delta E_p = mg\Delta h = 850\times9.81\times6.0$

Working in two steps: $850\times9.81=8338.5$ , then $8338.5\times6.0=50031\text{ J}$ .

Check by recomputing the second step differently: $8338.5\times6.0=8338.5\times5+8338.5=41692.5+8338.5=50031\text{ J}$ . Both routes agree.

So $\Delta E_p = \boxed{50031}\text{ J}\approx5.00\times10^{4}\text{ J}$ (3 s.f.).

Part (b): Gain in kinetic energy

The lift starts from rest and reaches $2.5\text{ m s}^{-1}$ , so the gain in kinetic energy equals the final kinetic energy: $\Delta E_k = \tfrac12mv^2 = \tfrac12\times850\times2.5^2$

Squaring the speed first: $2.5^2=6.25$ . Multiplying by the mass: $850\times6.25=5312.5$ . Halving: $\tfrac12\times5312.5=2656.25\text{ J}$ .

Check by recomputing in a different order: $\tfrac12\times850=425$ , and $425\times6.25=2656.25\text{ J}$ . Both routes agree.

So $\Delta E_k = \boxed{2656.25}\text{ J}\approx2.66\times10^{3}\text{ J}$ (3 s.f.).

Part (c): Total work done by the motor

Since no energy is lost to resistive forces, the principle of conservation of energy means the total work done by the motor is transferred entirely into the gain in gravitational PE and the gain in kinetic energy: $W_{\text{total}} = \Delta E_p + \Delta E_k = 50031 + 2656.25 = 52687.25\text{ J}$

So $W_{\text{total}} = \boxed{52687.25}\text{ J}\approx5.27\times10^{4}\text{ J}$ (3 s.f.).

Part (d): Average power output of the motor

Average power is the total work (energy transferred) divided by the time taken: $P_{\text{average}} = \frac{W_{\text{total}}}{t} = \frac{52687.25}{5.0}$

Dividing: $52687.25/5.0=10537.45\text{ W}$ .

Check: $10537.45\times5.0=52687.25\text{ J}$ , which matches $W_{\text{total}}$ from part (c). Confirmed.

So $P_{\text{average}} \approx\boxed{10537}\text{ W}\approx1.05\times10^{4}\text{ W}$ (3 s.f.).

Final answers

(a) Gain in gravitational PE $= \boxed{50031}\text{ J}\approx5.00\times10^{4}\text{ J}$
(b) Gain in kinetic energy $= \boxed{2656.25}\text{ J}\approx2.66\times10^{3}\text{ J}$
(c) Total work done by the motor $= \boxed{52687.25}\text{ J}\approx5.27\times10^{4}\text{ J}$
(d) Average power output $\approx\boxed{10537}\text{ W}\approx1.05\times10^{4}\text{ W}$