Q36E - University Physics with Modern Physics

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

Step-by-Step Solution

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 1: 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, $h=10 \mathrm{m}$.

Step 2: 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.

Step 3: (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 .

Step 4: (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.

Step 4: (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 .