Question

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 25.0° 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?

Answer 1

Answer:

-700j , 0j, 700j , 38.6j, 0j

Explanation:

from the question we are given given

distance (d) = 20m

frictional force (Ff) = 35N at 25 degree below the horizontal

(a) work done by friction = Ff x d

Ff is negative because from the question it is against the direction of motion

work done = -35 x 20 = 700 j

(b) work done by gravitational force = gravitational force x distance (cos θ) where  θ is the angle between the direction of motion ( which is on the x axis ) and the gravitational force (which is on the y axis)

the angle between the x and y axis ( direction of motion and gravitational force) is 90 degrees

cos 90 = 0, therefore work done by gravitational force is 0 j

(c) work done by the shopper = force applied by the shopper x distance

force applied by the shopper is of equal magnitude to the friction force and this is because the shopping cart is moving at constant speed.

therefore

work done by the shopper = 35 x 20 = 700 j

(d) force exerted by the shopper can be calculated from the formula

work done by the shopper = (force exerted by the shopper) x cos θ x distance

where cos θ is the  x component of the applied force

force exerted by the shopper becomes = work done / (distance x cos θ)

= 700 / (20 cos 25 ) = 38.6 N            

(e) net work done on the cart = net force x displacement

There is no net force because this cart is moving at constant speed and so the net work done is 0 j