Tag Archives: work done by a shopper to a cart

College Physics by Openstax Chapter 7 Problem 7

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 25.0º 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?

Solution:

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=Fdcos⁡θW=Fd \cos \theta

Part A. The Work Done on the Cart by Friction

In this case, the friction opposes the motion. So, we have the following given values:

F=35.0 Nd=20.0 mθ=180∘\begin{align*} F = & 35.0\ \text{N} \\ d = & 20.0\ \text{m} \\ \theta = & 180^{\circ } \\ \end{align*}
A shopper pusher a grocery cart showing that friction and displacement act in opposite directions.

The value of the angle θ\theta indicates that FF and dd are directed in opposite directions. Substituting these values into the formula,

W=Fdcos⁡θW=(35.0 N)(20.0 m)cos⁡180∘W=−700 N⋅mW=−700 J  (Answer)\begin{align*} W = & Fd \cos \theta \\ W = & \left( 35.0\ \text{N} \right)\left( 20.0\ \text{m} \right) \cos 180^{\circ } \\ W = & -700\ \text{N} \cdot \text{m} \\ W = & -700\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. Work Done on the Cart by the Gravitational Force

In this case, the gravitational force is directed downward while the displacement is horizontal as shown in the figure below.

A shopper pushes a grocery cart showing that displacement is horizontal while the gravitational force is downward.

We are given the following values:

F=mg d=20.0 mθ=90∘\begin{align*} F = & mg\ \\ d = & 20.0\ \text{m} \\ \theta = & 90^{\circ } \\ \end{align*}

Substituting these values into the work formula, we have

W=Fdcos⁡θW=(mg)(20.0 m)cos⁡90∘W=0 N⋅mW=0  (Answer)\begin{align*} W = & Fd \cos \theta \\ W = & \left( \text{mg} \right)\left( 20.0\ \text{m} \right) \cos 90^{\circ } \\ W = & 0\ \text{N} \cdot \text{m} \\ W = & 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

We can see that the gravitational force does not do any work on the cart because of the angle between the two quantities.

Part C. The Work on the Cart by the Shopper

Since we do not know the force exerted by the shopper, we are going to compute the work done by the shopper on the cart using the Work-Energy Theorem.

The work-energy theorem states that the net work WnetW_{\text{net}} on a system changes its kinetic energy. That is

Wnet=12mv2−12mv02W_{\text{net}} = \frac{1}{2}mv^{2}-\frac{1}{2}{mv_0} ^{2}

Now, we know that the shopper pushes the cart at a constant speed. This indicates that the initial and final velocities are equal to each other, making the net work WnetW_{\text{net}} is equal to zero.

Wnet=0W_{\text{net}} = 0

We also know that the total work done on the cart is the sum of the work done by the shopper and the friction force.

Wnet=Wshopper+Wfriction=0W_{\text{net}} = W_{\text{shopper}} +W_{\text{friction}}=0

This leaves us the final equation

Wshopper+Wfriction=0Wshopper+(−700 J)=0Wshopper=700 J  (Answer)\begin{align*} W_{\text{shopper}} + W_{\text{friction}} = & 0 \\ W_{\text{shopper}} + \left( -700\ \text{J} \right) = & 0 \\ W_{\text{shopper}} = & 700\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D. The force that the shopper exerts

In this case, the work of the shopper is directed 25 degrees below the horizontal while the displacement is still horizontal. This is depicted in the image below.

We are given the following values:

Wshopper=700 Jd=20.0 mθ=25∘\begin{align*} W_{\text{shopper}} = & 700\ \text{J} \\ d = & 20.0\ \text{m} \\ \theta = & 25^{\circ } \\ \end{align*}

Substituting these values in the formula for work, we have

Wshopper=Fshopperdcos⁡θFshopper=Wshopperdcos⁡θFshopper=700 J(20 m)cos⁡25∘Fshopper=38.6182 NFshopper=38.6 N  (Answer)\begin{align*} W_{\text{shopper}} & = F_{\text{shopper}} d \cos \theta \\ F_{\text{shopper}} & = \frac{W_{\text{shopper}}}{d \cos \theta} \\ F_{\text{shopper}} & = \frac{700\ \text{J}}{\left( 20\ \text{m} \right)\cos 25^{\circ}} \\ F_{\text{shopper}} & = 38.6182\ \text{N} \\ F_{\text{shopper}} & = 38.6\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part E. The Net Work done on the cart

The net work done on the cart is the sum of work done by each of the forces, namely friction and shopper forces. That is,

Wnet=Wshopper+WfrictionWnet=700 J+(−700 J)Wnet=0  (Answer)\begin{align*} W_{\text{net}} & = W_{\text{shopper}} + W_{\text{friction}} \\ W_{\text{net}} & = 700\ \text{J} + \left( -700\ \text{J} \right) \\ W_{\text{net}} & = 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}