Question

A shopper pushes a grocery cart for a distance 19 m at constant speed on level ground, against a 47.5 N frictional force. He pushes in a direction 22.5° below the horizontal.a) What is the work done on the cart by friction, in joules? b) What is the work done on the cart by the gravitational force, in joules? c) What is the work done on the cart by the shopper, in joules? d) Find the magnitude of the force, in newtons, that the shopper exerts on the cart.e) What is the total work done on the cart, in joules?

Answer 1

Given:

The frictional force on the cart is,

`f=47.5\text{ N}`

The direction of the push is

`\theta=22.5\degree\text{ below the horizontal}`

The grocery cart moves the distance

`d=19\text{ m}`

The cart moves with constant speed on level ground.

To find:

The work done on the cart by friction

The work done on the cart by the gravitational force

The work done on the cart by the shopper

The magnitude of the force that the shopper exerts on the cart.

The total work done on the cart, in joules

Step-by-step explanation:

a) The work by friction is,

`\begin{gathered} W=fd \\ =47.5*19 \\ =902.5\text{ J} \end{gathered}`

Hence, the work by friction is 902.5 J.

b) The free-body diagram is shown below:

The work by the gravitational force is zero as there is no vertical displacement of the cart.

c) The work done on the cart by the shopper is,

Force applied by the shopper in the direction of motion of the cart × The distance the cart moves

The net force along the level ground is zero as the cart is moving with constant speed. The force applied by the shopper is equal to the force of friction. The work by the shopper is,

`\begin{gathered} W_(shopper)=-W_(friction) \\ =-902.5\text{ J} \end{gathered}`

Hence, the work by the shopper is -902.5 J.

d) The force the shopper exerts, F, is given as follows

`\begin{gathered} Fcos\theta=f \\ F=(f)/(cos\theta) \\ F=(47.5)/(cos22.5\degree) \\ F=51.4\text{ N} \end{gathered}`

Hence, the force the shopper exerts on the cart is 51.4 N.

e) The total work done on the cart = The work done by the shopper+The work by the friction

The total work is,

`\begin{gathered} 902.5-902.5 \\ =0\text{ J} \end{gathered}`

Hence, the total work is zero on the cart.