Chapter 6, Problem 37QAP

To determine

(a)

The work done in the first 0.100 m by plotting a graph between F and x.

Answer to Problem 37QAP

The work done in the first 0.100 m is 0.812 J.

Explanation of Solution

Given:

The values of x and F.

x (m)	F (N)
0.0000	0.00
0.0100	2.00
0.0200	4.00
0.0300	6.00
0.0400	8.00
0.0500	10.00
0.0600	10.50
0.0700	11.00
0.0800	11.50
0.0900	12.00
0.1000	12.48
0.1100	12.48
0.1200	12.48
0.1300	12.60
0.1400	12.60
0.1500	12.70
0.1600	12.70
0.1700	12.60
0.1800	12.50
0.1900	12.50
0.2000	12.50
0.2100	12.48
0.2200	9.36
0.2300	6.24
0.2400	3.12
0.2500	0.00

Formula used:

Using an excel sheet, a graph is plotted between F and x and the work done is calculated by determining the area under the curve for the displacement under consideration.

Calculation:

Enter the values of x and F in an excel sheet and plot a graph as shown below.

For the displacement of 0.1 s, calculate the area under the curve.

The work done during the displacement of 0.1 s is equal to area OABC.

Calculate the area of the figure OABC.

  $\begin{array}{c}W=\text{ area }OABC\\ =\frac{1}{2}\left(OD\times AD\right)+\frac{1}{2}\left(AD+BC\right)\left(CD\right)\end{array}$

Substitute the values of the variables from the graph in the above equation.

  $\begin{array}{c}W=\frac{1}{2}\left(OD\times AD\right)+\frac{1}{2}\left(AD+BC\right)\left(CD\right)\\ =\frac{1}{2}\left[\left(0.050\text{ m}\right)\left(10.00\text{ N}\right)\right]+\frac{1}{2}\left[\left(10.00\text{ N}\right)+\left(12.48\text{ N}\right)\right]\left(0.050\text{ m}\right)\\ =0.812\text{ J}\end{array}$

Conclusion:

Thus, the work done in the first 0.100 m is 0.812 J.

To determine

(b)

The work done in the first 0.200 m by plotting a graph between F and x.

Answer to Problem 37QAP

The work done in the first 0.200 m is 2.06 J.

Explanation of Solution

Given:

The values of x and F.

x (m)	F (N)
0.0000	0.00
0.0100	2.00
0.0200	4.00
0.0300	6.00
0.0400	8.00
0.0500	10.00
0.0600	10.50
0.0700	11.00
0.0800	11.50
0.0900	12.00
0.1000	12.48
0.1100	12.48
0.1200	12.48
0.1300	12.60
0.1400	12.60
0.1500	12.70
0.1600	12.70
0.1700	12.60
0.1800	12.50
0.1900	12.50
0.2000	12.50
0.2100	12.48
0.2200	9.36
0.2300	6.24
0.2400	3.12
0.2500	0.00

Formula used:

Using an excel sheet, a graph is plotted between F and x and the work done is calculated by determining the area under the curve for the displacement under consideration.

Calculation:

The F-x graph is shown below:

The work done during the displacement of 0.200 m is given by the area OABEF.

Assume an average value of Force as 12.50 N during the displacement from 0.100 m to 0.0200 m.

  $\begin{array}{c}W=\text{ area }OABEF\\ =\frac{1}{2}\left(OD\times AD\right)+\frac{1}{2}\left(AD+BC\right)\left(CD\right)+\left(BE\times EF\right)\end{array}$

Substitute the values of the variables from the graph in the above equation.

  $\begin{array}{c}W=\frac{1}{2}\left(OD\times AD\right)+\frac{1}{2}\left(AD+BC\right)\left(CD\right)+\left(BE\times EF\right)\\ =\frac{1}{2}\left[\left(0.050\text{ m}\right)\left(10.00\text{ N}\right)\right]+\frac{1}{2}\left[\left(10.00\text{ N}\right)+\left(12.48\text{ N}\right)\right]\left(0.050\text{ m}\right)\\ +\left(0.100\text{ m}\right)\left(12.50\text{ N}\right)\\ =2.06\text{ J}\end{array}$

Conclusion:

Thus, the work done in the first 0.200 m is 2.06 J.

To determine

(c)

The work done during the displacement from 0.100 m to 0.200 m by plotting a graph between F and x.

Answer to Problem 37QAP

The work done during the displacement from 0.100 m to 0.200 m is 1.25 J.

Explanation of Solution

Given:

The values of x and F.

x (m)	F (N)
0.0000	0.00
0.0100	2.00
0.0200	4.00
0.0300	6.00
0.0400	8.00
0.0500	10.00
0.0600	10.50
0.0700	11.00
0.0800	11.50
0.0900	12.00
0.1000	12.48
0.1100	12.48
0.1200	12.48
0.1300	12.60
0.1400	12.60
0.1500	12.70
0.1600	12.70
0.1700	12.60
0.1800	12.50
0.1900	12.50
0.2000	12.50
0.2100	12.48
0.2200	9.36
0.2300	6.24
0.2400	3.12
0.2500	0.00

Formula used:

Using an excel sheet, a graph is plotted between F and x and the work done is calculated by determining the area under the curve for the displacement under consideration.

Calculation:

The F-x graph is shown below:

The work done during the displacement from 0.100 m to 0.200 m is given by the area BEFC.

Assume an average value of Force as 12.50 N during the displacement from 0.100 m to 0.0200 m.

  $\begin{array}{c}W=\text{ area }BEFC\\ =\left(BE\times EF\right)\end{array}$

Substitute the values of the variables from the graph in the above equation.

  $\begin{array}{c}W=\left(BE\times EF\right)\\ =\left(0.100\text{ m}\right)\left(12.50\text{ N}\right)\\ =1.25\text{ J}\end{array}$

Conclusion:

Thus, the work done during the displacement from 0.100 m to 0.200 m is 1.25 J.

To determine

(d)

The work done during the entire motion $0<x<0.250\text{ m}$.

Answer to Problem 37QAP

The work done during the entire motion $0<x<0.250\text{ m}$ is 2.37 J.

Explanation of Solution

Given:

The values of x and F.

x (m)	F (N)
0.0000	0.00
0.0100	2.00
0.0200	4.00
0.0300	6.00
0.0400	8.00
0.0500	10.00
0.0600	10.50
0.0700	11.00
0.0800	11.50
0.0900	12.00
0.1000	12.48
0.1100	12.48
0.1200	12.48
0.1300	12.60
0.1400	12.60
0.1500	12.70
0.1600	12.70
0.1700	12.60
0.1800	12.50
0.1900	12.50
0.2000	12.50
0.2100	12.48
0.2200	9.36
0.2300	6.24
0.2400	3.12
0.2500	0.00

Formula used:

Using an excel sheet, a graph is plotted between F and x and the work done is calculated by determining the area under the curve for the displacement under consideration.

Calculation:

The F-x graph is shown below:

The work done during the entire displacement is given by the area OABEG.

  $\begin{array}{c}W=\text{ area }OABEG\\ =\frac{1}{2}\left(OD\times AD\right)+\frac{1}{2}\left(AD+BC\right)\left(CD\right)+\left(BE\times EF\right)+\frac{1}{2}\left(FG\times EF\right)\end{array}$

Substitute the values of the variables from the graph in the above equation.

  $\begin{array}{c}W=\frac{1}{2}\left(OD\times AD\right)+\frac{1}{2}\left(AD+BC\right)\left(CD\right)+\left(BE\times EF\right)+\frac{1}{2}\left(FG\times EF\right)\\ =\frac{1}{2}\left[\left(0.050\text{ m}\right)\left(10.00\text{ N}\right)\right]+\frac{1}{2}\left[\left(10.00\text{ N}\right)+\left(12.48\text{ N}\right)\right]\left(0.050\text{ m}\right)\\ +\left(0.100\text{ m}\right)\left(12.50\text{ N}\right)+\frac{1}{2}\left(12.50\text{ N}\right)\left(0.050\text{ m}\right)\\ =2.37\text{ J}\end{array}$

Conclusion:

Thus, the work done during the entire motion $0<x<0.250\text{ m}$ is 2.37 J.