Chapter 6, Problem 13P

The force on a particle, acting along the x axis, varies as shown in Fig. 6-38. Determine the work done by this force to move the particle along the x axis: (a) from x=0.0 to x=10.0 m; (b) from x=0.0 to x=15.0 m.

To determine

The work done by the force to move the particle along the x-axis: (a) from x = 0 m to 10m

Answer to Problem 13P

Solution: 2800J

Explanation of Solution

Area under force-displacement graph gives work done by the force.

Given: A graph of force as a function of displacement.

Formula used: $\text{work done = area under F-x graph}$

Calculation: The work done by the force to move the particle along the x-axis from x = 0 to 10m is equal to the area of the curve in this interval.

  $\therefore \text{work done = area under F-x graph}$

  $\text{work done = area of the trapezium}$

  $\text{work done = }\frac{\text{1}}{\text{2}}\left(10+4\right)400=280\text{0}\text{J}$

Conclusion: The work done by the force to move the particle along the x-axis: from x = 0 to 10m is 2800J

To determine

The work done by the force to move the particle along the x-axis: from x = 0 to 15 m.

Answer to Problem 13P

Solution: 2100J

Explanation of Solution

Area under force-displacement graph gives work done by the force.

Given: A graph of force as a function of displacement.

Formula used: $\text{work done = area under F-x graph}$

Calculation: The work done by the force to move the particle along the x-axis from x = 10 to 15m is equal to the area of the curve in this interval.

  $\therefore \text{work done = area under F-x graph}$

  $\text{work done = area of the trapezium}$

  $\text{work done = }\frac{\text{1}}{\text{2}}\left(5+2\right)\left(-200\right)=-70\text{0}\text{J}$

The work done by the force to move the particle along the x-axis from x = 0 to 15m is

  $\text{work done = 2800 J-700 J=2100 J}$

Conclusion: The work done by the force to move the particle along the x-axis: from x = 0 to 15 m is 2100J.