A shopper in a supermarket pushes a cart with a force of 35 N directed at an ang

Question

A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 degree below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed. (a) Find the work done by the shopper as she moves down a 50.0-m length aisle. (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?

Explanation / Answer

a) The horizontal component of the applied force is 35 N x cos 25 = 31.72 N
Work = force x distance = 31.72 N x 50 m = 1,586 J

b) The work done on the cart is zero because there is no change in its mechanical energy.

c) If the applied force were horizontal, all of it would counter friction, so only 31.72 N would be needed. The reasoning of part b applies again.

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