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A 57.5kg Everlasting Gobstopper is lifted straight up to a distance of 20.0m by a cable attached to a motor and the job is done in 10.0s. Givens: Draw a FBD representing the direction of the influencing force, also note the direction of motion What is the force applied? What work is done? What power is developed by the motor?
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Free‑body diagram (describe): a dot for the gobstopper with two forces shown: tension T upward (from the cable) and weight mg downward. Direction of motion: upward (object is moved up 20.0 m).
Assume the lift is at constant speed (no net acceleration), so T = mg.
Calculations
- Force applied: T = mg = (57.5 kg)(9.80 m/s²) = 563.5 N (upward).
- Work done by the motor: W = T·Δy = mgΔy = 563.5 N · 20.0 m = 11 270 J.
- Average power: P = W / t = 11 270 J / 10.0 s = 1 127 W ≈ 1.13 kW.
(These values neglect any frictional losses or motor inefficiency.)
Assume the lift is at constant speed (no net acceleration), so T = mg.
Calculations
- Force applied: T = mg = (57.5 kg)(9.80 m/s²) = 563.5 N (upward).
- Work done by the motor: W = T·Δy = mgΔy = 563.5 N · 20.0 m = 11 270 J.
- Average power: P = W / t = 11 270 J / 10.0 s = 1 127 W ≈ 1.13 kW.
(These values neglect any frictional losses or motor inefficiency.)
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