Chapter 7, Problem 14P

To determine

The work done by the force is to be estimated.

Answer to Problem 14P

The work done by the force on the particle is 7.04J

Explanation of Solution

Given

It is given that the particle is moving along the axis with x=0 and x=4 and the graph is given below −

  

Formula used:

The following formula will be used to calculate the net work done by the force on the particle −

  $W_{net}=\Delta k$

Calculation:

The total area under the given curve of the graph is representing the work done by the force on the particle. It is represented as −

  

The final work done will be the summation of the area of triangle and rectangle under the curve −

Area of portion (1) = $=\frac{1}{2}\times 1N\times 1m=0.5J$

Area of portion 2(a) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion2\left(a\right)=1N\times 1m=1J$

Area of portion 2(b) = Area of triangle

  $Areaoftriangle=\frac{1}{2}\times b\times h$

  $Areaoftriangle=\frac{1}{2}\times \left(0.9N\right)\times 1m=0.45J$

Area of portion 2(c) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion2\left(c\right)=0.1N\times 1m=0.1J$

Area of portion 2(d) = Area of triangle

  $Areaoftriangle=\frac{1}{2}\times b\times h$

  $Areaoftriangle=\frac{1}{2}\times \left(0.1N\right)\times 0.1m=0.05J$

Area of portion 3(a) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion\text{ 3}\left(a\right)=1N\times 1m=1J$

Area of portion 3(b) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion\text{ 3}\left(b\right)=1N\times 1m=1J$

Area of portion 3(c) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion\text{ 3}\left(c\right)=1N\times 0.15m=0.15J$

Area of portion 3(d) = Area of triangle

  $Areaoftriangle=\frac{1}{2}\times b\times h$

  $Areaoftriangle=\frac{1}{2}\times \left(1N\right)\times 1.5m=0.75J$

Area of portion 4(a) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion\text{ 4}\left(a\right)=1N\times 1m=1J$

Area of portion 4(b) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion\text{ 4}\left(b\right)=0.6N\times 1m=0.6J$

Area of portion 4(c) = Area of rectangle

  $Areaofrectangle=base\times height$

  $Areaofportion\text{ 4}\left(c\right)=0.4N\times 0.6m=0.24J$

Area of portion 4(d) = Area of triangle

  $Areaoftriangle=\frac{1}{2}\times b\times h$

  $Areaoftriangle=\frac{1}{2}\times \left(0.4N\right)\times 0.4m=0.08J$

Area of portion 4(e) = Area of triangle

  $Areaoftriangle=\frac{1}{2}\times b\times h$

  $Areaoftriangle=\frac{1}{2}\times \left(0.6N\right)\times 0.4m=0.12J$

Now, the work done by the force on the particle can be given as −

  $W_{net}=\left(portion{\text{ 1}}_{area}+portion{\text{ 2}}_{area}+portion{\text{ 3}}_{area}+portion{\text{ 4}}_{area}\right)$

  $W_{net}=\left(\left(0.5\right)+\left(1+0.45+0.1+0.05\right)+\left(1+1+0.15+0.75\right)+\left(1+0.6+0.24+0.08+0.12\right)\right)$

  $W_{net}=7.04J$

Conclusion: It can be concluded that the work done by the force on particle is 7.04J.