Chapter 10, Problem 92A

(a)

To determine

The force exerted by the rope on the crate.

Answer to Problem 92A

The force exerted by the rope on the crate is $681\text{ N}$ .

Explanation of Solution

Given:

The work done is $W=11.4\text{ kJ}$ .

The displacement is $d=25.0\text{ m}$ .

The angle with the horizontal is $48.0°$

Formula used:

The expression for the work done is,

$W=Fd\cos \theta$

Here, $F$ is the force applied, $d$ is the displacement, and $\theta$ is the angle.

Calculation:

The force of the rope exert on the crate using the relation.

$\begin{array}{l}W=Fd\cos \theta \\ 11.4\text{ kJ}\times \frac{1,000\text{ J}}{1\text{ kJ}}=F\times \left(25.0\text{ m}\right)\times \cos 48.0°\\ F=681\text{ N}\end{array}$

Conclusion:

Thus, the force of the rope exert on the crate is $681\text{ N}$ .

(b)

To determine

The frictional force acting on the crate.

Answer to Problem 92A

The frictional force acting on the crate is $456\text{ N}$ .

Explanation of Solution

Given:

The work done is $W=11.4\text{ kJ}$ .

The displacement is $d=25.0\text{ m}$ .

The angle with the horizontal is $48.0°$

Calculation:

From part (a), the force of the rope exert on the crate is $F=681\text{ N}$ .

Consider the crate moves with constant speed, so the friction force equals to the horizontal component of the force of the rope.

The frictional force acting on the crate is,

$\begin{array}{l}{F}_{\text{f}}=F_{\text{x}}\\ F_{\text{f}}=F\cos \theta \\ F_{\text{f}}=\left(681\text{ N}\right)\cos 48.0°\\ F_{\text{f}}=456\text{ N}\end{array}$

Conclusion:

Thus, the frictional force acting on the crate is $456\text{ N}$ .

(c)

To determine

The energy transformed by the friction force between the floor and the crate.

Answer to Problem 92A

The energy transformed by the friction force between the floor and the crate is $-11.4\text{ kJ}$ .

Explanation of Solution

Given:

The work done is $W=11.4\text{ kJ}$ .

The displacement is $d=25.0\text{ m}$ .

The angle with the horizontal is $48.0°$

Formula used:

The expression for the work done is,

$W=Fd$

Here, $F$ is the force applied and $d$ is the displacement.

Calculation:

From part (b), the frictional force acting on the crate is $F_{\text{f}}=456\text{ N}$ .

The work done on the crate must be equal in magnitude and opposite in sign to the energy. The energy transformed by the friction force between the floor and the crate is,

$\begin{array}{l}W=-F_{\text{f}}d\\ W=-\left(456\text{ N}\right)\times \left(25.0\text{ m}\right)\\ W=-11.4\times {10}^3\text{ J}\times \frac{1\text{ kJ}}{1,000\text{ J}}\\ W=-11.4\text{ kJ}\end{array}$

Conclusion:

Thus, the energy transformed by the friction force between the floor and the crate is $W=-11.4\text{ kJ}$ .