Work and Power
4 marks · EasyA small elevator with total weight rises vertically in at a constant speed.
Calculate the work done by the elevator motor to raise the elevator.
State the increase in gravitational potential energy (GPE) of the elevator.
Calculate the power output of the elevator motor during the lift.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 7500 J
(b) 7500 J
(c) 375 W
Common mistakes: Confuses weight with mass; Incorrect formula for power
Show mark scheme and worked solution
Marking points
A1 (a) · 1 mark · Work done = 1500 N × 5 m = 7500 J.
B1 (b) · 1 mark · Increase in GPE = work done = 7500 J. (follow-through)
M1 (c) · 1 mark · Uses power = work / time.
A2 (c) · 1 mark · Calculates correctly: 7500 J / 20 s = 375 W. (follow-through)
Worked solution
a) Work = force × distance = 1500 N × 5 m = 7500 J.
b) GPE increases by the same amount: 7500 J.
c) Power = work / time = 7500 J ÷ 20 s = 375 W.