Magnitudes of Work
Person A pushes a mass constantly with some force. The mass moves on a rough surface with coefficent of friction . The mass moves a total of at constant velocity of while person A is pushing.
In a second experiment, a ball of mass falls from a height of . Person B catches it after falling , bringing it to rest.
Who does more work in total?
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1 solution
Although person A does not accelerate the mass, he must do work or the force of friction would remove energy from the mass. The force of friction is F f = μ m g = g Newtons, so the total work done by person A is W = F ⋅ d = g Joules in moving a meter.
The gravitational potential energy of the ball at the apex is U = m g h with respect to where the ball is caught. This potential energy is converted entirely kinetic energy which is then entirely removed due to the work of person B. Person B thus does work W = K = U = m g h = g Joules, the same amount of work.