Answer to Question #140053 in Physics for Ishpreet

Question

a shopper pushes a grocery cart 20.0m at constant speed on level ground against a 35.0N frictional force. He pushes in a direction 25.0 degrees 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 forces the shopper exerts, using energy considerations. (e) what is the total work done on the cart?

Answer 1

(a)

$$W_{\rm frict}=-F_{\rm frict}d=-35.0\times 20.0=-700\:\rm J$$

(b)

$$W_{\rm grav}=0\:\rm J$$

(c)

$$W_{\rm shopper}=-W_{\rm frict}=700\:\rm J$$

(d) Since a grocery cart moves at constant speed, so net force mut be zero

$$F\cos 25^{\circ}-F_{\rm frict}=0$$

The magnitude of force applied by a shopper

$$F=\frac{F_{\rm frict}}{\cos 25^{\circ}}=\frac{35.0\rm \: N}{\cos 25^{\circ}}=38.6\rm\:N$$

(e) Total work done on the cart is zero.