Chapter 9.2, Problem 48E

Work A car drives 500 ft on a road that is inclined 12° to the horizontal, as shown in the following figure. The car weighs 2500 lb. Thus gravity acts straight down on the car with a constant force F = −2500j. Find the work done by the car in overcoming gravity.

To determine

To find: The work done by the car in overcoming gravity.

Answer to Problem 48E

The work done by the car in overcoming gravity is 260,000 ft-lb.

Explanation of Solution

Given:

The distance covered by the car is 500 ft.

The angle formed by the car to the horizontal axis is $12°$.

The weight of car is 2500 lb.

A constant force F is $-2500j$.

Formula used:

The work done in moving an object along a vector D by a force F is,

$W=\left|F\right|\cdot \left|D\right|\cos \theta$ (1)

Where, $\theta$ is the angle between the force and displacement.

Calculation:

The Figure 1 which shown below the car exerts a force of 2500 lb in downward direction.

From Figure 1, it can be observed that the car exerts force into two components u and v.

The angle formed between u and w is,

$\begin{array}{c}\theta =90°-12°\\ =78°\end{array}$

The magnitude of force exerted by a car to overcome gravity is the component of w along v.

$\left|v\right|=\text{component}\text{of}w\text{along}v=w\cos \theta$

Substitute 2500 for w and $78°$ for $\theta$ in the above formula.

$\begin{array}{c}\left|v\right|=2500\cos 78°\\ =2500\times 0.207\left(\because \cos 78°=0.207\right)\\ =519.77\approx 520\text{lb}\end{array}$

From Figure 1, it can be noticed that the force is acting in the direction of movement.

Thus, the angle between force and displacement is 0.

Substitute 520 lb for F, 500 ft for D and 0 for $\theta$ in equation (1).

$\begin{array}{c}W=520\times 500\cos 0°\\ =260,000\text{ft-lb}\left(\because \cos 0°=1\right)\end{array}$

Thus, the work done by the car in overcoming gravity is 260,000 ft-lb.