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Work, Energy, and Power Example

Easy DifficultyFE Dynamics

A constant force acts on a body over a displacement. Find the work done by the force and the average power if the work is done in a given time.

Concept

Work done by a constant force F in the direction of displacement d is W = F \cdot d (units: joules, J). Kinetic energy of a particle is K E = \frac{1}{2} m v^{2} . Power is the rate of doing work: average power P = W / t or P = F \cdot v (units: watts, W). W = F \cdot d (constant force, same direction as displacement) K E = \frac{1}{2} m v^{2} P = \frac{W}{t} = F \cdot v (average power) Notation: W = work, F = force, d = displacement (in direction of force), K E = kinetic energy, P = power, t = time, v = speed.

A box is pushed along a horizontal floor by a constant horizontal force of 50 N. The box moves 8 m in the direction of the force, and this motion takes 5 seconds. Find: The work done by the force on the box. The average power delivered by the force during the 5 seconds.

F = 50 N (constant, horizontal) d = 8 m (displacement in direction of force) t = 5 s

W = F \cdot d (constant force, same direction as displacement) W = (50) (8) = 400 J

P = \frac{W}{t} = F \cdot v (average power) P = \frac{400}{5} = 80 W

(1) Work done: W = 400 J . (2) Average power: P = 80 W .

W = F d; K E = \frac{1}{2} m v^{2}; P = \frac{W}{t} = F v