A child applies a force $\overrightarrow{\mathbf{F}}$parallel to the x-axis to a 10.0kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applies varies with the x-coordinate of the sled as shown in Fig. E6.36. Calculate the work done by $\overrightarrow{\mathbf{F}}$when the sled moves

(a) from $\mathit{x}\mathbf{=}\mathbf{0}$to $\mathit{x}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$;

(b) from$\mathbf{x}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$to $\mathbf{x}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$;

(c) from$\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{m}$to $\mathbf{x}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$.

The given data can be listed below,

• The mass of the child is, $m=10\mathrm{kg}$.
• The maximum height of the triangle is, $h=10\mathrm{m}$.

If a body really moves a certain distance in the direction of the applied force, such movement is considered to constitute work. The dot product of force time distance is known as work.

When x rises during the displacement, the work done is positive is given by,

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

Here, b is the base of the triangle, and h is the height of the triangle.

Substitute all the values in the above from the graph,

$$\begin{gathered} W=\frac{1}{2}\left(8\mathrm{m}\right)\left(10\mathrm{N}\right) \\ =40\left(\mathrm{N}.\mathrm{m}\right)\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =40\mathrm{J} \end{gathered}$$

Thus, the work done by the force on the sled when it moves from$x=0$ to$x=8\mathrm{m}$ is 40 J .

When x rises during the displacement, the work done is positive is given by,

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

Here, b is the base of the triangle, and h is the height of the triangle.

Substitute all the values in the above from the graph,

$$\begin{gathered} W=\frac{1}{2}\left(4\mathrm{m}\right)\left(10\mathrm{N}\right) \\ =20\left(\mathrm{N}·\mathrm{m}\right)\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =20\mathrm{J} \end{gathered}$$

Thus, the work done by the force on the sled when it moves from$x=8\mathrm{m}$ to$x=12\mathrm{m}$ is 20 J.

When x rises during the displacement, the work done is positive is given by,

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

Here, b is the base of the triangle, and h is the height of the triangle.

Substitute all the values in the above from the graph,

$$\begin{gathered} W=\frac{1}{2}\left(12\mathrm{m}\right)\left(10\mathrm{N}\right) \\ =60\left(\mathrm{N}·\mathrm{m}\right)\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =60\mathrm{J} \end{gathered}$$

Thus, the work done by the force on the sled when it moves from $\mathrm{x}=0\mathrm{m}$to $\mathrm{x}=12\mathrm{m}$is 60 J .