Chapter 7, Problem 1P

A shopper in a supermarket pushes a cart with a force of 35.0 N directed at an angle of 25.0° 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 50.0-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?

To determine

The work done by the shopper on the cart.

Answer to Problem 1P

The work done by the shopper on the cart is $\text{1}\text{.5}\times {\text{10}}^3\text{kJ}$.

Explanation of Solution

Given Information:

The force exerted by the shopper on the cart is $35.0\text{N}$ and the inclination of the aisle below the horizontal is $25.0°$. The length of the aisle is $50.0\text{m}$.

Formula to calculate the work done by the shopper is,

$W=F\Delta r\cos \theta$

• $W$ is the work done by the shopper.
• $F$ is the force exerted by the shopper on the cart.
• $\Delta r$ is the distance through which cart moves.
• $\theta$ is the inclination angle of the aisle from the horizontal.

Substitute $35.0\text{N}$ for $F$, $25.0°$ for $\theta$ and $50.0\text{m}$ for $\Delta r$ to find $W$.

$\begin{array}{c}W=\left(35.0\text{N}\right)\left(50.0\text{m}\right)\cos \left(25.0°\right)\\ =1586\text{N}\cdot \text{m}\times \frac{{10}^{-3}\text{kJ}}{1\text{N}\cdot \text{m}}\\ =1.5\times {10}^3\text{kJ}\end{array}$

Conclusion:

Therefore, the work done by the shopper on the cart is $\text{1}\text{.5}\times {\text{10}}^3\text{kJ}$.

To determine

Whether the force applied by the shopper will be larger, smaller or same by pushing the cart horizontally at same speed.

Introduction: The work done by the force is equal to the dot product of the applied force vector to the displacement vector of the body.

Explanation of Solution

Given Information:

The force exerted by the shopper on the cart is $35.0\text{N}$ and the inclination of the aisle below the horizontal is $25.0°$. The length of the aisle is $50.0\text{m}$.

Formula to calculate the work done by the shopper on the cart is,

$W=F\Delta r\cos \theta$

Rearrange the above equation,

$F=\frac{W}{\Delta r\cos \theta }$ (I)

Since the shoppers moves with the constant velocity therefore the distance travel by the shopper is same. Also the change in kinetic energy of the cart is zero that corresponds to work done must be same. So the applied force by the shopper is depends upon the slope of the aisle.

The value of $\cos 0°>\cos 25.0°$ hence from the equation (I) the applied force by the shopper on the cart in horizontal case must be smaller in compare to the force applied by the shopper in the part (a).

Conclusion:

Therefore, the applied force by the shopper will be smaller in compare to the applied force on the cart in part (a).

To determine

Whether the work done by the shopper will change or remains same by pushing the cart horizontally at same speed.

Introduction: The work done by the force is equal to the dot product of the applied force vector to the displacement vector of the body.

Explanation of Solution

Given Information:

The force exerted by the shopper on the cart is $35.0\text{N}$ and the inclination of the aisle below the horizontal is $25.0°$. The length of the aisle is $50.0\text{m}$.

From the work energy theorem, the work done by the shopper on the cart is equal to the change in kinetic energy of the cart. Since the cart is moving with the same speed so, its kinetic energy remains same and hence the work done by the shopper on the cart remains same as that of part (a).

Conclusion:

Therefore, the work done by the shopper to push the cart horizontally remains same. It does not change.