Q6-43E - University Physics with Modern Physics

Question

A force $\overrightarrow{F}$ is applied to a $2.0\text{\hspace{0.17em}kg}$, radio-controlled model car parallel to the x-axis as it moves along a straight track. The x-component of the force varies with the x-coordinate of the car (Fig. E6.43). Calculate the work done by the force $\overrightarrow{F}$ when the car moves from

(a)  $x=0\text{\hspace{0.17em}to\hspace{0.17em}}x=3.0\text{\hspace{0.17em}m}$

(b)  $x=3\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=4.0\text{\hspace{0.17em}m}$

(c)  $x=4.0\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=7.0\text{\hspace{0.17em}m}$

(d)  $x=0\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=7.0\text{\hspace{0.17em}m}$

(e) $x=7.0\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=2.0\text{\hspace{0.17em}m}$

Step-by-Step Solution

Answer

a) The work done by the force when the car moves from $x=0\text{\hspace{0.17em}to\hspace{0.17em}}x=3.0\text{\hspace{0.17em}m}$ is $4.0\text{\hspace{0.17em}J}$.

b) The work done by the force when the car moves from $x=3\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=4.0\text{\hspace{0.17em}m}$ is $0\text{\hspace{0.17em}J}$.

c) The work done by the force when the car moves from $x=4\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=7.0\text{\hspace{0.17em}m}$ is $-1.0\text{\hspace{0.17em}J}$.

d) The work done by the force when the car moves from $x=0\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=7.0\text{\hspace{0.17em}m}$ is $\text{3.0\hspace{0.17em}J}$.

e) The work done by the force when the car moves from is $x=7.0\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=2.0\text{\hspace{0.17em}m}$ $-1.0\text{\hspace{0.17em}J}$.

Step 1: Identification of given data

The given data can be listed below,

• The mass of the model car is, $m=2\text{\hspace{0.17em}kg}$.
• The x-component of force varies with the x-axis from the graph.

Step 2: Concept/Significance of force

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.

Step 3: (a) Determination of the work done by the force when F → the car moves from x = 0  to  x = 3 .0  m

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,

 $W_1=\frac{1}{2}(b\times h)+l\times w$

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,

$\begin{array}{c}{W}_1=\frac{1}{2}(2\text{\hspace{0.17em}N}\times 2.0\text{\hspace{0.17em}m})+2.0\text{\hspace{0.17em}N}\times 1.0\text{\hspace{0.17em}m}\\ =4.0\text{\hspace{0.17em}J}\end{array}$

 

Thus, the work done by the force when the car moves from $x=0\text{\hspace{0.17em}to\hspace{0.17em}}x=3.0\text{\hspace{0.17em}m}$  is $4.0\text{\hspace{0.17em}J}$.

Step 4: (b) Determination of the work done by the force F → when the car moves from x = 0  to  x = 3 .0  m

As the displacement between $3\text{\hspace{0.17em}m}$ to $4\text{\hspace{0.17em}m}$ 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  $x=3\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=4.0\text{\hspace{0.17em}m}$ is $0\text{\hspace{0.17em}J}$

Step 5: (c) Determination of the work done by the force F → when the car moves from x = 4 .0  m to  x = 7 .0  m

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,

 $W_3=-\frac{1}{2}(b\times h)$

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,

$\begin{array}{c}{W}_3=-\frac{1}{2}(1.0\text{\hspace{0.17em}N}\times 2.0\text{\hspace{0.17em}m})\\ =-1.0\text{\hspace{0.17em}J}\end{array}$

Thus,the work done by the force when the car moves from $x=4\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=7.0\text{\hspace{0.17em}m}$ is $-1.0\text{\hspace{0.17em}J}$.

Step 6: (d) Determination of the work done by the force F → when the car moves from x = 0  m to  x = 7 .0  m

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,$x=0\text{\hspace{0.17em}to\hspace{0.17em}}x=3.0\text{\hspace{0.17em}m}$,$W_2$  is the work done on the car when moves from $x=3\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=4.0\text{\hspace{0.17em}m}$, and $W_3$  is the work done on the car when moves from $x=4\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=7.0\text{\hspace{0.17em}m}$.

 

Substitute all the values in the above,

 $\begin{array}{c}{W}_4=4.0\text{\hspace{0.17em}J}+0\text{\hspace{0.17em}J}-1.0\text{\hspace{0.17em}J}\\ =\text{3.0\hspace{0.17em}J}\end{array}$

Thus, the work done by the force when the car moves from $x=0\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=7.0\text{\hspace{0.17em}m}$  is $\text{3.0\hspace{0.17em}J}$.

Step 3: (e) Determination of the work done by the force F → when the car moves from x = 7 .0  m to  x = 2 .0  m

The work done by the force on the model car between $7\text{\hspace{0.17em}m}$ to $3\text{\hspace{0.17em}m}$ is positive $1.0\text{\hspace{0.17em}J}$ as the force and displacement are both in negative x-direction, and the work done by the force on the model car between $3\text{\hspace{0.17em}m}$ to $2\text{\hspace{0.17em}m}$ is negative $2.0\text{\hspace{0.17em}J}$ as the force on the car is in positive x- direction and the displacement is in negative x-direction, so the work done for $7.0\text{\hspace{0.17em}m}$ to $2\text{\hspace{0.17em}m}$ is given by,

 $\begin{array}{c}{W}_5=1.0\text{\hspace{0.17em}J}+(-2.0\text{\hspace{0.17em}J})\\ =-1.0\text{\hspace{0.17em}J}\end{array}$

Thus, the work done by the force when the car moves from $x=7.0\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}}x=2.0\text{\hspace{0.17em}m}$ is $-1.0\text{\hspace{0.17em}J}$.