91Ó°ÊÓ

The given data can be listed below,

• The initial position of the sled is,$x=0$.
• The mass of the child is$m=10\text{ k²µ}$,.
• The maximum height of the triangle is$h=10\text{″¾}$,.

Work Energy Theorem focuses on how an item's kinetic energy changes as a result of the work performed by the net force exerted on the object.

From the triangle, the work done on the/of the sled is given by,

$W=\frac{1}{2}b×h$

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,

$\begin{array}{c}W=\frac{1}{2}\left(8\text{″¾}\right)\left(10\text{ N}\right)\\ =40(\text{N}â‹\dots \text{m})\left(\frac{1\text{ J}}{1\text{ N}â‹\dots \text{m}}\right)\\ =40\text{ J}\end{array}$

According to the work energy theorem, the work done on the/of the sled is given by,

$\begin{array}{c}{W}_{tot}=K_f−K_i\\ =\frac{1}{2}m{v}_f^2−\frac{1}{2}m{v}_i^2\end{array}$

Here, m is the mass of the sled and $v_f$is the final velocity of the sled,$v_1$ is the initial velocity of the sled.

Substitute all the values in the above the velocity of the sled is,

$\begin{array}{c}40\text{ J}=\frac{1}{2}m{v}_f^2−0\\ v_f=\sqrt{\frac{80\text{ J}}{10\text{ k²µ}}}\\ =2.83\text{″¾/s}\end{array}$

Thus, the velocity of the of sled at$x=8\text{″¾}$ is $2.83\text{″¾/s}$.

From the triangle, the work done on the/ of the sled is given by,

$W=\frac{1}{2}b×h$

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,

$\begin{array}{c}W=\frac{1}{2}\left(12\text{″¾}\right)\left(10\text{ N}\right)\\ =60(\text{N}â‹\dots \text{m})\left(\frac{1\text{ J}}{1\text{ N}â‹\dots \text{m}}\right)\\ =60\text{ J}\end{array}$

Also, from the work energy theorem, the work done on the / of the sled is given by,

$\begin{array}{c}{W}_{tot}=K_f−K_i\\ =\frac{1}{2}m{v}_f^2−\frac{1}{2}m{v}_i^2\end{array}$

Here, m is the mass of the sled and $v_f$is the final velocity of the sled,$v_1$ is the initial velocity of the sled.

Substitute all the values in the above the velocity of the sled is,

$\begin{array}{c}60\text{ J}=\frac{1}{2}m{v}_f^2−0\\ v_f=\sqrt{\frac{2×60\text{ J}}{10\text{ k²µ}}}\\ =3.46\text{″¾/s}\end{array}$

Thus, the velocity of the of the sled at $x=12\text{″¾}$is$3.46\text{″¾/s}$ .