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 W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=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:

\begin{align*}
F = & 35.0\ \text{N} \\
d = & 20.0\ \text{m} \\
\theta = & 180^{\circ } \\
\end{align*}

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

\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.

We are given the following values:

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

Substituting these values into the work formula, we have

\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 W_{\text{net}} on a system changes its kinetic energy. That is

W_{\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 W_{\text{net}} is equal to zero.

W_{\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.

W_{\text{net}} = W_{\text{shopper}} +W_{\text{friction}}=0

This leaves us the final equation

\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:

\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

\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,

\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*}