Chapter 9, Problem 8PQ

A 537-kg trailer is hitched to a truck. Find the work done by the truck on the trailer in each of the following cases. Assume rolling friction is negligible. a. The trailer is pulled at constant speed along a level road for 2.30 km. b. The trailer is accelerated from rest to a speed of 88.8 km/h. c. The trailer is pulled at constant speed along a road inclined at 12.5° for 2.30 km.

To determine

The work done for by the truck on the trailer when the trailer is pulled at constant speed.

Answer to Problem 8PQ

The work done for by the truck on the trailer when the trailer is pulled at constant speed is $\underset{\_}{0}$.

Explanation of Solution

Write the equation of work done.

  $W_{tot}=K_f-K_i$                                                                                                    (I)

Here, $W_{tot}$ is the work done, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy.

At constant speed, the initial and final kinetic energy is same.

  $K_f=K_i$                                                                                                              (II)

Conclusion:

Substitute, $K_f$ for $K_i$ in equation (I) to find $W_{tot}$.

  $\begin{array}{c}{W}_{tot}=K_f-K_i\\ =0\end{array}$

Thus, the work done for by the truck on the trailer when the trailer is pulled at constant speed is $\underset{\_}{0}$.

To determine

The work done for by the truck on the trailer when the trailer is pulled at 88.8km/h.

Answer to Problem 8PQ

The work done for by the truck on the trailer when the trailer is pulled at 88.8km/h is $\underset{\_}{1.64\times {10}^5\text{J}}$.

Explanation of Solution

Write the equation of work done.

  $W_{tot}=K_f-K_i$                                                                                                  (III)

Here, $W_{tot}$ is the work done, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy.

In this case, the initial kinetic energy is zero.

  $K_i=0$                                                                                                              (IV)

Write the expression for the final kinetic energy.

  $K_f=\frac{1}{2}m{v}^2$                                                                                                        (V)

Here, $m$ is the mass and $v$ is the speed.

Rewrite the expression from equation (I).

  $W_{tot}=\frac{1}{2}m{v}^2$                                                                                                    (VI)

Conclusion:

Substitute $537\text{kg}$ for $m$ and $88.8\text{km/h}$ for $v$ in equation (VI) to find $W_{tot}$.

  $\begin{array}{c}{W}_{tot}=\frac{1}{2}\left(537\text{kg}\right)\left[\left(88.8\text{km/h}\right)\left(\frac{0.277778\text{m/s}}{1\text{km/h}}\right)\right]\\ =\left(268.5\text{kg}\right)\left[\left(88.8\text{km/h}\right)\left(\frac{0.277778\text{m/s}}{1\text{km/h}}\right)\right]\\ =1.64\times {10}^5\text{J}\end{array}$

Thus, the work done for by the truck on the trailer when the trailer is pulled at 88.8km/h is $\underset{\_}{1.64\times {10}^5\text{J}}$.

To determine

The work done for by the truck on the trailer when the trailer is pulled at an angle $12.5°$.

Answer to Problem 8PQ

The work done for by the truck on the trailer when the trailer is pulled at an angle $12.5°$ is $\underset{\_}{2.62\times {10}^6\text{J}}$.

Explanation of Solution

Write the equation of work done.

  $W_T=-W_g$                                                                                                 (VII)

Here, $W_T$ is the work done by the truck and $W_g$ is the work done by gravity.

Write the expression for the work done by gravity.

  $W_g=mgr\cos \left(90+\theta \right)$                                                                                     (VIII)

Here, $m$ is the mass $g$ is the gravitational acceleration, $r$ is the distance and $\theta$ is the angle.

Rewrite the expression from equation (VIII).

  $W_T=-mgr\cos \left(90+\theta \right)$                                                                                 (IX)

Conclusion:

Substitute $537\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $2.30\text{km}$ for $r$ and $12.5°$ for $\theta$ in equation (IX) to find $W_T$.

  $\begin{array}{c}{W}_T=-\left(537\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left[\left(2.30\text{km}\right)\left(\frac{1\times {10}^3\text{m}}{1\text{km}}\right)\right]\cos \left(90°+\left(12.5°\right)\right)\\ =-\left(537\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(2.30\times {10}^3\text{m}\right)\cos \left(102.5°\right)\\ =2.62\times {10}^6\text{J}\end{array}$

Thus, the work done for by the truck on the trailer when the trailer is pulled at an angle $12.5°$ is $\underset{\_}{2.62\times {10}^6\text{J}}$.