Chapter 6, Problem 28P

(a)

To determine

The work done by the force from $x=0.0$ to $x=2.0\text{m}$ .

Explanation of Solution

Given:

Mass of the object is $3.0\text{kg}$ .

Velocity of the object is $2.4\text{m}/\text{s}$ .

Formula used:

Write expression for work done by the force from $x=0.0$ to $x=2.0\text{m}$ .

  $W=nA$  ........ (1)

Here, $n$ is number of squares and $A$ is area of one square.

Write expression for area of square.

  $A=xy$  ........ (2)

Here, $x$ is value of distance covered in one unit and $y$ is value of force applied.

Calculation:

Substitute $0.25\text{m}$ for $x$ and $0.5\text{N}$ for $y$ in equation (2).

  $\begin{array}{l}A=\left(0.25\text{m}\right)\left(0.5\text{N}\right)\\ A=0.125\text{J}\end{array}$

Substitute $22$ for $n$ and $0.125\text{J}$ for $A$ in equation (1).

  $\begin{array}{l}W=\left(22\right)\left(0.125\text{J}\right)\\ W=2.75\text{J}\end{array}$

Conclusion:

Thus, work done by the force is $2.75\text{J}$ .

(b)

To determine

The kinetic energy of object at $x=2.0\text{m}$ .

Explanation of Solution

Given:

Mass of the object is $3.0\text{kg}$ .

Velocity of the object is $2.4\text{m}/\text{s}$ .

Formula used:

Write expression for work done by the force from $x=0.0$ to $x=2.0\text{m}$ .

  $W=nA$  ........ (1)

Here, $n$ is number of squares and $A$ is area of one square.

Write expression for area of square.

  $A=xy$  ........ (2)

Here, $x$ is value of distance covered in one unit and $y$ is value of force applied.

Write expression for kinetic energy at $x=2.0\text{m}$ .

  $K_{2\text{m}}=K_0+W$

Substitute $\frac{1}{2}m{v}^2$ for $K_0$ in above expression.

  $K_{2\text{m}}=\frac{1}{2}m{v}^2+W$  ........ (3)

Calculation:

Substitute $0.25\text{m}$ for $x$ and $0.5\text{N}$ for $y$ in equation (2).

  $\begin{array}{l}A=\left(0.25\text{m}\right)\left(0.5\text{N}\right)\\ A=0.125\text{J}\end{array}$

Substitute $22$ for $n$ and $0.125\text{J}$ for $A$ in equation (1).

  $\begin{array}{l}W=\left(22\right)\left(0.125\text{J}\right)\\ W=2.75\text{J}\end{array}$

Substitute $3.0\text{kg}$ for $m$ , $2.4\text{m}/{\text{s}}^{\text{2}}$ and $2.75\text{J}$ in equation (3).

  $\begin{array}{l}{K}_{2\text{m}}=\frac{1}{2}\left(3.0\text{kg}\right){\left(2.4\text{m}/{\text{s}}^{\text{2}}\right)}^2+2.75\text{J}\\ K_{2m}=11.4\text{J}\end{array}$

Conclusion:

Thus, kinetic energy at $x=2.0\text{m}$ is $11.4\text{J}$ .

(c)

To determine

The speed of object at $x=2.0\text{m}$ .

Explanation of Solution

Given:

Mass of the object is $3.0\text{kg}$ .

Velocity of the object is $2.4\text{m}/\text{s}$ .

Formula used:

  $K_{2\text{m}}=\frac{1}{2}m{v}^2+W$

Write expression for velocity at $x=2.0\text{m}$ .

  $v_{2\text{m}}=\sqrt{\frac{2K_{2\text{m}}}{m}}$

Calculation:

Substitute $11.4\text{J}$ for $K_{2\text{m}}$ and $3.0\text{kg}$ for $m$

  $\begin{array}{l}{v}_{2\text{m}}=\sqrt{\frac{2\left(11.4\text{J}\right)}{3.0\text{kg}}}\\ v_{2\text{m}}=2.8\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of object is $2.8\text{m}/\text{s}$ .

(d)

To determine

The work done on the object from $x=0.0\text{m}$ to $x=4.0\text{m}$ .

Explanation of Solution

Given:

Mass of the object is $3.0\text{kg}$ .

Velocity of the object is $2.4\text{m}/\text{s}$ .

Formula used:

Write expression for work done on the object from $x=0.0\text{m}$ to $x=4.0\text{m}$ .

  $W=nA$  ........ (1)

Here, $n$ is number of squares and $A$ is area of one square.

Write expression for area of square.

  $A=xy$  ........ (2)

Here, $x$ is value of distance covered in one unit and $y$ is value of force applied.

Calculation:

Substitute $0.25\text{m}$ for $x$ and $0.5\text{N}$ for $y$ in equation (2).

  $\begin{array}{l}A=\left(0.25\text{m}\right)\left(0.5\text{N}\right)\\ A=0.125\text{J}\end{array}$

Substitute $26$ for $n$ and $0.125\text{J}$ for $A$ in equation (1).

  $\begin{array}{l}W=\left(26\right)\left(0.125\text{J}\right)\\ W=3.25\text{J}\end{array}$

Conclusion:

Thus, work done is $3.25\text{J}$ .

(e)

To determine

The speed of object at $x=4.0\text{m}$ .

Explanation of Solution

Given:

Mass of the object is $3.0\text{kg}$ .

Velocity of the object is $2.4\text{m}/\text{s}$ .

Formula used:

Write expression for kinetic energy at $x=4.0\text{m}$ .

  $K_{\text{4m}}=K_0+W$

Substitute $\frac{1}{2}m{v}_{4\text{m}}^2$ for $K_0$ in above expression.

  $K_{\text{4m}}=\frac{1}{2}m{v}_{4\text{m}}^2+W$  ........ (1)

Write expression for velocity at $x=4.0\text{m}$ .

  $v_{\text{4m}}=\sqrt{\frac{2K_{\text{4m}}}{m}}$  ........ (2)

Calculation:

Substitute $3.0\text{kg}$ for $m$ , $2.4\text{m}/{\text{s}}^{\text{2}}$ and $3.25\text{J}$ in equation (3).

  $\begin{array}{l}{K}_{\text{4m}}=\frac{1}{2}\left(3.0\text{kg}\right){\left(2.4\text{m}/{\text{s}}^{\text{2}}\right)}^2+3.25\text{J}\\ K_{4m}=11.9\text{J}\end{array}$

Substitute $11.4\text{J}$ for $K_{2\text{m}}$ and $3.0\text{kg}$ for $m$ in equation (4).

  $\begin{array}{l}{v}_{\text{4m}}=\sqrt{\frac{2\left(11.9\text{J}\right)}{3.0\text{kg}}}\\ v_{\text{4m}}=2.8\text{m}/\text{s}\end{array}$

Conclusion:

Thus, velocity of the object is $2.8\text{m}/\text{s}$ .