Chapter 7, Problem 10SP

A motor furnishes 120 hp to a device that lifts a 5000-kg load to a height of 13.0 m in a time of 20 s. Find the efficiency of the machine.

To determine

The efficiency of a machine, considering the motor delivers a power of $120\text{ hp}$ to a device, which lifts a $5000\text{ kg}$ load to a height of $13.0\text{ m}$ in $20\text{ s}$.

Answer to Problem 10SP

Solution:

$36\%$

Explanation of Solution

Given data:

The power delivered by the motor to the device is $120\text{ hp}$.

The mass of the lifted load is $5000\text{ kg}$.

The height to which the load was lifted is $13.0\text{ m}$.

The time taken to lift the load to the height is $20\text{ s}$.

Formula used:

The formula for efficiency of a machine is

$\eta =\frac{P_o}{P_i}\times 100\%$

Here, $\eta$ is the efficiency of the machine;$P_o$ and $P_i$ represent the output power and input power of the machine, respectively.

The work done is calculated by the formula:

$w=\left(F\right)\left(S\right)$

Here, $w$ is the work done, $F$ is the force, and $S$ is the displacement in the direction of force $F$.

Power is calculated by the formula:

$P=\frac{w}{t}$

Here, $P$ is the power and $t$ is the time taken to do the work $w$.

The formula for weight of a body is

$W=mg$

Here, $W$ is the weight of the body, $m$ is the mass of the body, and $g$ is the acceleration due to gravity.

Explanation:

Consider the expression for weight of the lifted load as

$W=mg$

The standard value of $g$ is $9.81{\text{ m/s}}^2$.

Substitute $5000\text{kg}$ for $m$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}W=\left(5000\text{kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =49,050\text{ N}\end{array}$

Consider the expression for work done to lift the load to a height of $13.0\text{ m}$ as

$w_0=Wh$

Here, $h$ is the height to which the load is lifted and $w_0$ is the corresponding work done.

Substitute $49050\text{ N}$ for $F$ and $13.0\text{ m}$ for $S$

$\begin{array}{c}{w}_0=\left(49,050\text{ N}\right)\left(13.0\text{ m}\right)\\ =637,650\text{ J}\end{array}$

Calculate the power delivered by the machine by the formula:

$P_o=\frac{w_0}{t}$

Here, $t$ is the time taken to lift the load to the height of $13.0\text{ m}$.

Substitute $20\text{ s}$ for $t$ and $637650\text{ J}$ for $w_0$

$\begin{array}{c}{P}_o=\frac{637650\text{ J}}{20\text{ s}}\\ =31882.5\text{ W}\\ =\left(31882.5\text{ W}\right)\left(\frac{1\text{hp}}{745.7\text{W}}\right)\\ =42.76\text{ hp}\end{array}$

Consider the expression for efficiency of a machine:

$\eta =\frac{P_o}{P_i}\times 100\%$

Substitute $42.76\text{ hp}$ for $P_o$ and $120\text{ hp}$ for $P_i$

$\begin{array}{c}\eta =\frac{42.76\text{ hp}}{120\text{ hp}}\times 100\%\\ \simeq 36\%\end{array}$

Conclusion:

The efficiency of the machine is $36\%$.