Chapter 10, Problem 75A

To determine

The input work required to lift a $215\text{kg}$ mass by a distance of $5.65\text{m}$ using a machine that is $72.5\%$ efficient.

Answer to Problem 75A

  $1.64\times {10}^4\text{J}$

Explanation of Solution

Given:

The efficiency of the machined, $e=72.5\%$

The mass to be lifted, $m=215\text{kg}$

The distance by which the mass is to be lifted, $d=5.65\text{m}$

Formula used:

Work done is given by,

  $W=Fd\cos \theta$

Where $F$ is the force done, $d$ is the distance, and $\theta$ is the angle between the distance and force.

And, the efficiency if given by,

  $e=\frac{W_o}{W_i}\times 100$

Where $W_o$ is the output work done, and $W_i$ is the input work done.

Calculation:

Now, the weight of the mass that is to be lifted is,

  $F=mg=215\left(9.8\right)=2107\text{N}$

Thus, the force required to pull the mass is $F=2107\text{N}$

And, the angle between this force and the displacement is $\theta =0°$

Thus, the output work done is given by,

  $\begin{array}{c}{W}_o=Fd\cos \theta \\ =\left(2107\right)\left(5.65\right)\cos 0°\\ =1.19\times {10}^4\text{J}\end{array}$

Then, the efficiency of the given lever can be written as,

  $\begin{array}{c}e=\frac{W_o}{W_i}\times 100\\ 72.5=\frac{1.19\times {10}^4}{W_i}\times 100\\ W_i=1.64\times {10}^4\text{J}\end{array}$

Conclusion:

Thus, the input work required is $1.64\times {10}^4\text{J}$ .