Chapter 10, Problem 54A

a.

To determine

Work done by escalator on SL.

Answer to Problem 54A

  $5.96\times {10}^4\text{J}$

Explanation of Solution

Given:

Mass of SL is $m=52\text{kg}$ .

Length of escalator is $d=227\text{m}$ .

Average inclination is $\theta =31°$ .

Formula used:

Work done is given by,

  $W=\Delta E=\Delta PE=mgh$    …..    (1)

Where, $\Delta E$ is change in energy, $\Delta PE$ is change in potential energy, $m$ is mass, $g$ is gravitationalacceleration and $h$ is height gained.

Calculation:

From the given condition the figure is,

  

From the figure, height $h$ gained is:

  $\begin{array}{l}\sin \theta =\frac{h}{d}\\ \Rightarrow h=d\sin \theta \end{array}$

And $g=9.8{\text{m/s}}^2$

Putting the values in (1)

  $\begin{array}{c}W=mgd\sin \theta \\ =52\times 9.8\times 227\sin 31°\\ =5.96\times {10}^4\text{J}\end{array}$

Conclusion:

So, work done by escalator on SL is $W=5.96\times {10}^4\text{J}$ .

b.

To determine

Work done by escalator on P.

Answer to Problem 54A

Work done by escalator on P is $7.44\times {10}^4\text{J}$

Explanation of Solution

Given:

Mass of Pis $m=65\text{kg}$ .

Length of escalator is $d=227\text{m}$ .

Average inclination is $\theta =31°$ .

Formula used:

Work done is given by,

  $W=\Delta E=\Delta PE=mgh$    …..    (2)

Where $\Delta E$ is change in energy, $\Delta PE$ is change in potential energy, $m$ is mass, $g$ is gravitationalacceleration and $h$ is height gained.

Calculation:

From the given condition the figure is,

  

Now, from the figure height $h$ gained is,

  $\begin{array}{l}\sin \theta =\frac{h}{d}\\ \Rightarrow h=d\sin \theta \end{array}$

And $g=9.8{\text{m/s}}^2$

Putting the valuesin (2)

  $\begin{array}{c}W=mgd\sin \theta \\ =65\times 9.8\times 227\sin 31°\\ =7.44\times {10}^4\text{J}\end{array}$

Conclusion:

So, work done by escalator on P is $W=7.44\times {10}^4\text{J}$ .