Q44E - University Physics with Modern Physics

Question

Suppose the $\mathbf{2}\mathbf{ }\mathbf{\text{kg}}$ model car in Exercise 6.43 is initially at rest at $\mathit{x}\mathbf{=}\mathbf{0}$ and $\overrightarrow{\mathbf{F}}$is the net force acting on it. Use the work-energy theorem to find the speed of the car at (a) $\mathit{x}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{\text{m}}$; (b) $\mathit{x}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{\text{m}}$; (c) $\mathit{x}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{\text{m}}$.

Step-by-Step Solution

Answer

1. The velocity of the car at $\text{x}=\text{3} \text{m}$ is $2 \text{m/s}$.
2. The velocity of the car at $\text{x}=4 \text{m}$ is $2 \text{m/s}$.
3. The velocity of the car at $\text{x}=7 \text{m}$ is $1.73 \text{m/s}$
   

Step 1: Identification of given data

The given data can be listed below,

 

The mass of the car is, $\text{m}=\text{2} \text{kg}$

The initial position of the car is, $\text{x}=\text{0} \text{m}$

Step 2: Concept/Significance of velocity.

The pace at which an object's location changes in relation to a point of reference is known as its velocity, which is a time-dependent quantity. A statement of its speed and direction of motion is what is meant by velocity.

Step 3: (a) Determination of the find the speed of the car at x = 3   m

The work done due to force is also positive which is given by,

${\text{W}}_{\text{1}}=\frac{\text{1}}{\text{2}}\left(\text{b}\times \text{h}\right)+\text{l}\times \text{w}$

Here, from graph b is the base of the triangle 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,

$$\begin{gathered} {\text{W}}_{\text{1}}=\frac{1}{2}\left(2 \text{N}\times 2.0 \text{m}\right)+2.0 \text{N}\times 1.0 \text{m} \\ =4.0 \text{J} \end{gathered}$$

According to the work-energy theorem, the work done on the car is given by,

$\begin{array}{rcl}\text{W}& =& {\text{K}}_{\text{f}}-{\text{K}}_{\text{i}}\\ & =& \frac{\text{1}}{\text{2}}{\text{mv}}_{\text{f}}^{\text{2}}-\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{i}}^{\text{2}}\end{array}$

Here, m is the mass of the car,  ${\text{v}}_{\text{f}}$ is the final velocity of the car, and ${\text{v}}_{\text{i}}$  is the initial velocity of the car.

Substitute all the values in above for the velocity of the car.

$$\begin{gathered} 4 \text{J}=\frac{1}{2}\left(2 \text{kg}\right){\text{v}}_{\text{1f}}^{\text{2}}-0 \\ {\text{v}}_{\text{1f}}=\sqrt{\frac{2\times 4 \text{J}}{2 \text{kg}}} \\ {\text{v}}_{\text{1f}}=2 \text{m/s} \end{gathered}$$

Thus, the velocity of the car at $\text{x}=\text{3} \text{m}$ is $2 \text{m/s}$

Step 4: (b) Determination of the find the speed of the car at x = 4   m

The work done from $\text{x}=\text{0} \text{m} \text{to} \text{x}=\text{4} \text{m}$is same as the work done for data-custom-editor="chemistry" $\text{x}=3 \text{m}$ is same. So the velocity of the car at $\text{x}=4 \text{m}$is given by,

$$\begin{gathered} \text{W}=\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{2f}}^{\text{2}} \\ {\text{v}}_{\text{2f}}=\sqrt{\frac{\text{2W}}{\text{m}}} \end{gathered}$$

Here, W is the work done on the car at $\text{x}=4 \text{m}$ and m is the mass of the car.

 

Substitute all the values in the above,

$\begin{array}{rcl}{\text{v}}_{\text{2f}}& =& \sqrt{\frac{2\times 4 \text{J}}{2 \text{kg}}\left(\frac{1 \text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1 \text{J}}\right)}\\ & =& \sqrt{4} \text{m/s}\\ & =& 2 \text{m/s}\end{array}$

Thus, the velocity of the car at $\text{x}=4 \text{m}$is $2 \text{m/s}$.

Step 5: (c) Determination of the find speed of the car at x = 7   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,

$\text{W}={\text{W}}_{\text{1}}+{\text{W}}_{\text{3}}$

Here, $W_1$is the work done on the car when moves from$\text{x}=0 \text{m} \text{to} \text{x}=3.0 \text{m}$ and ${\text{W}}_{\text{3}}$ is the work done on the car when moves from $\text{x}=4 \text{m} \text{to} \text{x}=7.0 \text{m}$

Substitute all the values in the above,

$\begin{array}{rcl}\text{W}& =& 4.0 \text{J}-1.0 \text{J}\\ & =& \text{3.0} \text{J}\end{array}$

The velocity of the car at $\text{x}=7 \text{m}$ is given by,

$$\begin{gathered} \text{W}=\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{f}}^{\text{2}} \\ {\text{v}}_{\text{f}}=\sqrt{\frac{\text{2W}}{\text{m}}} \end{gathered}$$

Substitute all the values in the above,

$\begin{array}{rcl}{\text{v}}_{\text{f}}& =& \sqrt{\frac{\text{2W}}{\text{m}}}\\ & =& \sqrt{\frac{2\times 3 \text{J}}{2 \text{kg}}}\left(\frac{1 \text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1 \text{J}}\right)\\ & =& \sqrt{3} \text{m/s}\\ & =& 1.73 \text{m/s}\end{array}$

Thus, the velocity of the car at $\text{x}=7 \text{m}$ is $1.73 \text{m/s}$.