Question

A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in Fig.7-39. The scale of the figure’s vertical axis is set by${\mathbf{F}}_{\mathbf{s}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. How much work is done by the force as the block moves from the origin to $\mathbf{x}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$?

Short answer

The net work done by the force using a graph $\mathrm{W}=25\mathrm{J}$.

Step-by-step solution

Given

1. The mass of block is, $\mathrm{m}=5.0\mathrm{kg}$
2. The scale for vertical axis is, ${\mathrm{F}}_{\mathrm{s}}=10.0\mathrm{N}$

Understanding the concept

We can use the concept that the work done is equal to the area under the curve for a graph.

Formula:

$$\begin{gathered} \mathbf{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}\mathbf{F}\left(x\right)\mathbf{d}\mathbf{x} \\ \mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathbf{base}\mathbf{\times }\mathbf{height} \\ \mathbf{A}\mathbf{=}\mathbf{length}\mathbf{\times }\mathbf{breadth} \end{gathered}$$

Calculate the net work done

We know that the total area will be the net work done by the force.

So,

$\mathrm{W}={\mathrm{W}}_1+{\mathrm{W}}_2+{\mathrm{W}}_3+{\mathrm{W}}_4$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{W}=10\mathrm{N}\times 2\mathrm{m}+\frac{1}{2}\times 10\mathrm{N}\times 2\mathrm{m}+0+\left(-\frac{1}{2}\times 2\mathrm{N}\times 5\mathrm{m}\right) \\ \mathrm{W}=25\mathrm{J} \end{gathered}$$

Therefore, the net work done is 25J