91Ó°ÊÓ

The given data can be listed below,

• The mass of the model car is, $m=2\text{ k²µ}$.
• The x-component of force varies with the x-axis from the graph.

Force is defined as the acceleration or change in position of a massed object caused by the application of external energy or power. Considering that a force is a vector quantity, it has both a magnitude and a direction.

The force acting on the model car and the displacement are both positive so the work done due to force is also positive, which is given by,

$W_1=\frac{1}{2}(b×h)+l×w$

Here, from graph b is the base of the triangle (starts from 0 till 2 m), h is the height of the triangle, l is the length of the rectangle (from 3 m to 4 m), and w is the width of the rectangle.

Substitute all the values in the above,

$\begin{array}{c}{W}_1=\frac{1}{2}(2\text{ N}×2.0\text{″¾})+2.0\text{ N}×1.0\text{″¾}\\ =4.0\text{ J}\end{array}$

Thus, the work done by the force when the car moves from$x=0\text{ t´Ç }x=3.0\text{″¾}$ is $4.0\text{ J}$.

As the displacement between $3\text{″¾}$to $4\text{″¾}$is zero, so the force applied on the model car is also zero; due to this, there is no work is done.

Thus, the work done by the force when the car moves from $x=3\text{″¾â€‰t´Ç }x=4.0\text{″¾}$is$0\text{ J}$

The force acting on the car in the x-direction is negative, and displacement is positive, so the work done will also be negative, which can be calculated as,

$W_3=−\frac{1}{2}(b×h)$

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

Substitute all the values in the above,

$\begin{array}{c}{W}_3=−\frac{1}{2}(1.0\text{ N}×2.0\text{″¾})\\ =−1.0\text{ J}\end{array}$

Thus,the work done by the force when the car moves from$x=4\text{″¾â€‰t´Ç }x=7.0\text{″¾}$ is $−1.0\text{ J}$.

The work done due to the force acting on the car is the sum of all the work done within the range of 0 to 7 m given by,

$W_4=W_1+W_2+W_3$

Here,$W_1$ is the work done on the car when moves from,$x=0\text{ t´Ç }x=3.0\text{″¾}$,$W_2$ is the work done on the car when moves from $x=3\text{″¾â€‰t´Ç }x=4.0\text{″¾}$, and$W_3$ is the work done on the car when moves from $x=4\text{″¾â€‰t´Ç }x=7.0\text{″¾}$.

Substitute all the values in the above,

$\begin{array}{c}{W}_4=4.0\text{ J}+0\text{ J}−1.0\text{ J}\\ =\text{3.0 J}\end{array}$

Thus, the work done by the force when the car moves from$x=0\text{″¾â€‰t´Ç }x=7.0\text{″¾}$ is $\text{3.0 J}$.

The work done by the force on the model car between $7\text{″¾}$to $3\text{″¾}$is positive $1.0\text{ J}$as the force and displacement are both in negative x-direction, and the work done by the force on the model car between $3\text{″¾}$to $2\text{″¾}$is negative $2.0\text{ J}$as the force on the car is in positive x- direction and the displacement is in negative x-direction, so the work done for $7.0\text{″¾}$to $2\text{″¾}$is given by,

$\begin{array}{c}{W}_5=1.0\text{ J}+(−2.0\text{ J})\\ =−1.0\text{ J}\end{array}$

Thus, the work done by the force when the car moves from $x=7.0\text{″¾â€‰t´Ç }x=2.0\text{″¾}$is $−1.0\text{ J}$.