Question

Assignments

1. 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

Hello there!

I hope you and your family are staying safe and healthy during this unprecendented time.

A) What is the work done?

Answer: We need to use the formula

`w=-F_f(d)`

`w=-(35)(20)`

`w=-700J`

B) What is the work done on the cart by the gravitational force?

Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:

`w=Fdcos90`

`w=0`

C) What is the work done on the cart by the shopper?

This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force

`W shopper + W friction = 0\\W shopper = W-friction \\W shopper = 700J`

D) Find the force the shopper exerts, using energy considerations.

`F_f+Fcos25=0\\-35+Fcos25=0\\F=38.6N`

E) What is the total work done?

You just need to add them:

`w=wshopper+wfriction\\w=0`