Chapter 10, Problem 94A

(a)

To determine

The power developed by the winch.

Answer to Problem 94A

The power developed by the winch is $3.1\times {10}^2\text{ W}$ .

Explanation of Solution

Given:

The force applied on the crate is $F=875\text{ N}$ .

The angle with the horizontal is $\theta =15°$

The speed is $v=0.25\text{m}/\text{s}$ .

The coefficient of friction between the crate and the incline is $\mu =0.45$ .

Formula used:

The expression for the work done is

  $W=Fd$

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

The expression for the power is

  $P=\frac{W}{t}$

Here, $t$ is the time period.

The expression for the velocity is

  $v=\frac{d}{t}$

Calculation:

The work of the crate is done by the winch, gravity, and friction. The kinetic energy of the crate does not change, so the sum of the three works is equal to zero.

The normal force on the crate as follows:

  $\begin{array}{l}{F}_N=F\cos \theta \\ F_N=\left(875\text{ N}\right)\cos 15°\\ F_N=845.2\text{ N}\end{array}$

The power developed by the winch as follows:

  $\begin{array}{l}{P}_{\text{winch}}=P_{\text{friction}}+P_{\text{gravity}}\\ P_{\text{winch}}=\frac{W_{\text{friction}}}{t}+\frac{W_{\text{gravity}}}{t}\\ P_{\text{winch}}=\frac{F_{\text{friction}}d}{t}+\frac{F_{\text{gravity}}d}{t}\\ P_{\text{winch}}=\mu F_{N}v+Fv\end{array}$

  $\begin{array}{l}{P}_{\text{winch}}=0.45\times \left(845.2\text{ N}\right)\times \left(0.25\text{m}/\text{s}\right)+\left(875\text{ N}\right)\times \left(0.25\text{m}/\text{s}\right)\\ P_{\text{winch}}=3.1\times {10}^2\text{ W}\end{array}$

Conclusion:

Therefore, the power developed by the winch is $3.1\times {10}^2\text{ W}$ .

(b)

To determine

The electrical power delivered to the winch.

Answer to Problem 94A

The electrical power delivered to the winch is $3.6\times {10}^2\text{ W}$ .

Explanation of Solution

Given:

The force applied on the crate is $F=875\text{ N}$ .

The angle with the horizontal is $\theta =15°$

The speed is $v=0.25\text{m}/\text{s}$ .

The coefficient of friction between the crate and the incline is $\mu =0.45$ .

The efficiency of the winch is $e=85\%$ .

Formula used:

The expression for efficiency is

  $e=\frac{P_{\text{o}}}{P_{\text{i}}}$

Here, $P_{\text{o}}$ is the power output and $P_{\text{i}}$ is the power input.

Calculation:

From part (a), the power developed by the winch is $P_{\text{o}}=3.1\times {10}^2\text{ W}$ .

The electrical power delivered to the winch as follows:

  $\begin{array}{l}e=\frac{P_{\text{o}}}{P_{\text{i}}}\\ \frac{85}{100}=\frac{3.1\times {10}^2\text{ W}}{P_{\text{i}}}\\ P_{\text{i}}=3.6\times {10}^2\text{ W}\end{array}$

Conclusion:

Therefore, the electrical power delivered to the winch is $3.6\times {10}^2\text{ W}$ .