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 the 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:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance =  20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.