Question

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 $\mathbf{x}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$.

Short answer

1. The work done by the force on the sled when it moves from $x=0$to $x=8\mathrm{m}$is $40\mathrm{J}$.
2. 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 .
3. The work done by the force on the sled when it moves from $x=0\mathrm{m}$to $x=12\mathrm{m}$is 60 J.

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of work done.

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.

(a) Determination of the work done by F→ when the sled moves from x=0 to x=8.0 m

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 .

(b) Determination of the work done by F→ when the sled moves from x=0 m to x=12.0 m

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.

(c) Determination of the work done by F→ when the sled moves from x=0 m to x=12.0 m

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 .