Problem 7PE

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0º below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart?

Step 1 of 5</p>

The man pushes the cart in a direction below the horizontal at a constant speed against a friction force.

a) We need to find out the work done on the cart by friction.

The work done by the friction is in the opposite direction of motion.

So the work done by friction is given by

The angle

So

F is the frictional force

d is the distance,

Step 2 of 5</p>

b) We need to find out the work done on the cart by the gravitational force.

Work done by the gravitational force is perpendicular to direction of motion so the value of .

So the work done on the cart by the gravitational force is

Step 3 of 5</p>

c) We need to find out the work done by the shopper on the cart

By work energy theorem we can say the net work done on the cart is equal to the sum of work done by the shopper and the work done by the friction is equal to zero.