Question

A shopper in a supermarket pushes a cart with a force of $\mathbf{35}\mathit{N}$ directed at an angle of $\mathbf{25}\mathbf{°}$ below the horizontal. The force is just sufficient to balance various friction forces, so the cart moves at constant speed.

(a) Find the work done by the shopper on the cart as she moves down a $\mathbf{50}\mathit{m}$-long aisle.

(b) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the friction force doesn’t change, would the shopper’s applied force be larger, smaller, or the same?

(c) What about the work done on the cart by the shopper?

Short answer

(a) The work done by the shopper when the force is directed along the aisle is .$\mathbf{1750}\mathit{J}$

(b) To maintain the same speed, the shopper applies larger force than before.

(c) The work done on the cart by the shopper remains the same.

Step-by-step solution

Given data

Force applied in the first case

$F_1=35\text{\hspace{0.17em}N}$

Angle made by the force with the horizontal

${\theta }_h=25°$

Distance traveled along the slop

$s=50\text{\hspace{0.17em}m}$

Work done

Work done by a force $F$ causing displacement $s$ at an angle $\theta$ with the force is

$\mathit{W}\mathbf{=}\mathit{F}\mathit{s}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$ .....(I)

Determining work done in the first case

In the first case the force is parallel to the displacement, so the angle between them is zero. From equation (I), the work done is

$\begin{array}{c}W=35\text{\hspace{0.17em}N}\times 50\text{\hspace{0.17em}m}\times \cos 0\\ =1750\cdot (1\text{\hspace{0.17em}N}\cdot \text{m}\times \frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}N}\cdot \text{m}})\\ =1750\text{\hspace{0.17em}J}\end{array}$

Thus the work done is .$1750\text{\hspace{0.17em}J}$

Determining the force applied in the second case

The force along the aisle will have to be the same to cancel out the friction. Let the horizontal force applied be $F_h$. Then

$\begin{array}{c}{F}_h\cos {\theta }_h=35\text{\hspace{0.17em}N}\\ F_h=\frac{35\text{\hspace{0.17em}N}}{\cos 25°}\\ =38.62\text{\hspace{0.17em}N}\end{array}$

Thus the applied force is .$38.62\text{\hspace{0.17em}N}$

Determining the work done in the second case

Since the force along the aisle remains the same, the work done will remain the same.