Chapter 9, Problem 42A

What change in kinetic energy does an airplane experience on takeoff if it is moved a distance of 500 m by a sustained net force of 5000 N?

To determine

Change in kinetic energy.

Answer to Problem 42A

  $\Delta KE=2.5\times {10}^6\text{J}$

Explanation of Solution

Given:

Net force, $F=5000\text{N}$

Distance moved, $d=500\text{m}$

Formula Used:

The work done is given as,

  $W=Fd$

Where, $F$ is force and $d$ is displacement.

The change in kinetic energy is equal work done that is

Δ KE = Work

Calculation:

On substituting the values, the work done is

  $\begin{array}{l}W=5000\times 500\\ =2.5\times {10}^6\text{J}\end{array}$

As the change in kinetic energy is equal work done so,

  $\Delta KE=W=2.5\times {10}^6\text{J}$

Conclusion:

Therefore, the change in kinetic energy of airplane is $\Delta KE=2.5\times {10}^6\text{J}$