Chapter 10, Problem 61A

a.

To determine

Work done to pull the object.

Answer to Problem 61A

  $255\text{J}$

Explanation of Solution

Given:

Displacement is 7 meters.

Formula Used:

The area under force vs displacement graph gives the work done.

Area of triangle: $\frac{1}{2}\text{base×height}$

Area of rectangle: $\text{length×width}$

Calculation:

From the graph:

From 0 to 2 m:

It is a triangle of height 20 m and base 2 m. So,

  $\begin{array}{c}{W}_1=\frac{1}{2}\times 20\times 2\\ =20\text{J}\end{array}$

From 2 to 3 m:

It is a triangle of base 1 m and height 30 m and a rectangle of length 20 m and width 1 m. So,

  $\begin{array}{c}{W}_2=\left(\frac{1}{2}\times 30\times 1\right)+\left(20\times 1\right)\\ =35\text{J}\end{array}$

From 3 to 7 m:

It is a rectangle of length 50 m and width 4 m. So,

  $\begin{array}{c}{W}_3=50\times 4\\ =200\text{J}\end{array}$

Hence total work is,

  $\begin{array}{c}W=W_1+W_2+W_3\\ =20+35+200\\ =255\text{J}\end{array}$

Conclusion:

The total work done to pull the object is 255 J.

b.

To determine

Power developed.

Answer to Problem 61A

  $127.5\text{W}$

Explanation of Solution

Given:

Displacement is 7 meters.

Time given is $t=2\text{s}$

Formula used:

Power developed is given as,

  $P=\frac{W}{t}$

Where, $P$ is power developed, $W$ is work done and $t$ is time taken to do work.

Calculation:

As work done for 7 meters is $255\text{J}$ .

So, power developed for $2\text{s}$ will be

  $\begin{array}{c}P=\frac{W}{t}\\ =\frac{255}{2}\\ =127.5\text{W}\end{array}$

Conclusion:

Power developed in 2 s is 127.5 W.