Question

A shopper in a supermarket pushes a cart with a force of 40 N directed at an angle of 25° 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 35.0-m length aisle. ____ Joules

(b) What is the net work done on the cart? ____ Joules

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?

Answer 1

A )
Work = Force * distance
Since the distance is horizontal, you need the horizontal component of the force.
Horizontal component = 40 * cos 25˚

Work done by shopper = 40 * cos 25˚ * 35


B ) The friction force is horizontal.
Since the force is just sufficient to balance various friction forces, so the cart moves at constant speed, the magnitude of the total friction force must equal the magnitude of the horizontal component of the force of the shopper. The direction of the total friction force is opposite the direction of motion.

So the friction force = -40 * cos 25˚

The net force on the cart = 40 * cos 25˚ + -40 * cos 25˚ = 0
The net work done on the cart by all forces = 0 * 35 = 0

C) If the applied force were horizontal, all of it would counter friction, so only 40 * cos 25˚ N would be needed. The reasoning of part b applies again.