Concept explainers
(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.
Substitute the values of the variables from the graph in the above equation.
Conclusion:
Thus, the work done in the first 0.100 m is 0.812 J.
(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.
Substitute the values of the variables from the graph in the above equation.
Conclusion:
Thus, the work done in the first 0.200 m is 2.06 J.
(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.
Substitute the values of the variables from the graph in the above equation.
Conclusion:
Thus, the work done during the displacement from 0.100 m to 0.200 m is 1.25 J.
(d)
The work done during the entire motion
Answer to Problem 37QAP
The work done during the entire motion
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.
Substitute the values of the variables from the graph in the above equation.
Conclusion:
Thus, the work done during the entire motion
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Chapter 6 Solutions
COLLEGE PHYSICS
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