Chapter 10, Problem 91A

(a)

To determine

The work done by the mover.

Answer to Problem 91A

The work done by the mover is $1.04\times {10}^3\text{ J}$ .

Explanation of Solution

Given:

The force applied on the dolly is $F=496\text{ N}$ .

The mass of the refrigerator is $m=115\text{ kg}$

The distance of the ramp is $d_{\text{ramp}}=2.10\text{ m}$ .

The rise of the ramp is $d=0.850\text{ m}$ .

Formula used:

The expression for the work done is,

$W=Fd$

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

Calculation:

The work done by the mover is,

$\begin{array}{l}{W}_{\text{i}}=F{d}_{\text{ramp}}\\ W_{\text{i}}=\left(496\text{ N}\right)\times \left(2.10\text{ m}\right)\\ W_{\text{i}}=1.04\times {10}^3\text{ J}\end{array}$

Conclusion:

Thus, the work done by the mover is $1.04\times {10}^3\text{ J}$ .

(b)

To determine

The work done on the refrigerator by the machine.

Answer to Problem 91A

The work done on the refrigerator by the machine is $958\text{ J}$ .

Explanation of Solution

Given:

The force applied on the dolly is $F=496\text{ N}$ .

The mass of the refrigerator is $m=115\text{ kg}$

The distance of the ramp is $d_{\text{ramp}}=2.10\text{ m}$ .

The rise of the ramp is $d=0.850\text{ m}$ .

Formula used:

The expression for the work done is,

$W=mgd$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $d$ is the distance.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

The work done on the refrigerator by the machine is,

$\begin{array}{l}{W}_{\text{o}}=mgd\\ W_{\text{o}}=\left(\text{115 kg}\right)\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\times \left(\text{0}\text{.850 m}\right)\\ W_{\text{o}}=958\text{ J}\end{array}$

Conclusion:

Thus, the work done on the refrigerator by the machine is $958\text{ J}$ .

(c)

To determine

The efficiency of the machine.

Answer to Problem 91A

The efficiency of the machine is $92.1\%$ .

Explanation of Solution

Given:

The force applied on the dolly is $F=496\text{ N}$ .

The mass of the refrigerator is $m=115\text{ kg}$

The distance of the ramp is $d_{\text{ramp}}=2.10\text{ m}$ .

The rise of the ramp is $d=0.850\text{ m}$ .

Formula used:

The expression for efficiency is

  $e=\frac{W_{\text{o}}}{W_{\text{i}}}\times 100\%$

Here, $W_{\text{o}}$ is the work done by the machine and $W_{\text{i}}$ is the work done by the mover.

Calculation:

From part (a), the work done by the mover is $W_{\text{i}}=1.04\times {10}^3\text{ J}$ .

From part (b), the work done on the refrigerator by the machine is $W_{\text{i}}=958\text{ J}$ .

The efficiency of the machine is,

$\begin{array}{l}e=\frac{W_{\text{o}}}{W_{\text{i}}}\times 100\%\\ e=\frac{\text{958 J}}{1.04\times {10}^3\text{ J}}\times 100\%\\ e=92.1\%\end{array}$

Conclusion:

Thus, the efficiency of the machine is $92.1\%$ .