Chapter 11, Problem 45A

(a)

To determine

The work done on the train by the jet engine.

Answer to Problem 45A

The work done on the train by the jet engine is $2.55\times {10}^8\text{ J}$ .

Explanation of Solution

Given:

The mass of the train is $m=2.50\times {10}^4\text{ kg}$

The force applied on the train is $F=5.0\times {10}^5\text{ N}$ .

The distance of the level track is $d=509\text{ m}$ .

Formula used:

The expression for the work done is,

$W=Fd$

Here, $F$ is the force applied and $d$ is the displacement.

Calculation:

The work done on the train by the jet engine as follows:

$\begin{array}{l}W=Fd\\ W=\left(\text{5}\text{.0}\times {\text{10}}^5\text{ N}\right)\times \left(\text{509 m}\right)\\ W=2.55\times {10}^8\text{ J}\end{array}$

Conclusion:

Thus, the work done on the train by the jet engine is $2.55\times {10}^8\text{ J}$ .

(b)

To determine

The change in kinetic energy.

Answer to Problem 45A

The change in kinetic energy is $2.55\times {10}^8\text{ J}$ .

Explanation of Solution

Given:

The mass of the train is $m=2.50\times {10}^4\text{ kg}$

The force applied on the train is $F=5.0\times {10}^5\text{ N}$ .

The distance of the level track is $d=509\text{ m}$ .

Formula used:

The expression for the change in kinetic energy is,

$\Delta KE=W$

Here, $W$ is the work done.

Calculation:

From part (a), the work done on the train by the jet engine is $W=2.55\times {10}^8\text{ J}$ .

The change in kinetic energy is,

$\begin{array}{l}\Delta KE=W\\ \Delta KE=2.55\times {10}^8\text{ J}\end{array}$

Conclusion:

Thus, the change in kinetic energy is $2.55\times {10}^8\text{ J}$ .

(c)

To determine

The final kinetic energy of the train.

Answer to Problem 45A

The final kinetic energy of the train if it started from rest is $2.55\times {10}^8\text{ J}$ .

Explanation of Solution

Given:

The mass of the train is $m=2.50\times {10}^4\text{ kg}$

The force applied on the train is $F=5.0\times {10}^5\text{ N}$ .

The distance of the level track is $d=509\text{ m}$ .

The train is initially started from rest.

Formula used:

Change in kinetic energy is the difference between the final kinetic energy and initial kinetic energy.

Calculation:

From part (b), the change in kinetic energy is $\Delta KE=2.55\times {10}^8\text{ J}$ .

Consider the initial kinetic energy $K{E}_{\text{i}}=0$ . Since, the train is started from rest.

The final kinetic energy of the train if it started from rest is,

$\begin{array}{l}\Delta KE=K{E}_{\text{f}}-K{E}_{\text{i}}\\ 2.55\times {10}^8\text{ J}=K{E}_{\text{f}}-0.0\text{ J}\\ K{E}_{\text{f}}=2.55\times {10}^8\text{ J}\end{array}$

Conclusion:

Thus, the final kinetic energy of the train if it started from rest is $2.55\times {10}^8\text{ J}$ .

(d)

To determine

The final speed of the train.

Answer to Problem 45A

The final speed of the train is $143\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the train is $m=2.50\times {10}^4\text{ kg}$

The force applied on the train is $F=5.0\times {10}^5\text{ N}$ .

The distance of the level track is $d=509\text{ m}$ .

Formula used:

The expression for the kinetic energy is,

$KE=\frac{1}{2}m{v}^2$

Here, $m$ is the mass and $v$ is the speed.

Calculation:

From part (b), the final kinetic energy of the train is $K{E}_{\text{f}}=2.55\times {10}^8\text{ J}$ .

The speed of the train is,

$\begin{array}{l}K{E}_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}^2\\ v_{\text{f}}=\sqrt{\frac{2K{E}_{\text{f}}}{m}}\\ v_{\text{f}}=\sqrt{\frac{2\times \left(2.55\times {10}^8\text{ J}\right)}{\left(2.50\times {10}^4\text{ kg}\right)}}\\ v_{\text{f}}=143\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of the train is $143\text{m}/\text{s}$ .