# The initial kinetic energy imparted to a 0.0300 kg bullet is 1091 J. Assuming it accelerated down...

## Question:

The initial kinetic energy imparted to a 0.0300 kg bullet is 1091 J. Assuming it accelerated down a 1.00 m long rifle barrel, estimate the average power delivered to it during the firing. Answer in kW.

## Mechanical Power:

Consider a diesel locomotive moving on the track and the work done by the locomotive in time duration (t) is (W ). In this case, the power delivered by the locomotive can be given as,

{eq}P = \frac{W}{t}. {/eq}

The mechanical is power generally expressed in Watts (W) or horsepower (hp).

## Answer and Explanation:

Given data:

• Mass of the bullet, {eq}m = 0.0300 \ kg {/eq}
• Kinetic energy, {eq}K = 1091 \ J {/eq}
• Length of the barrel, {eq}d = 1.00 \ m {/eq}

Let the speed of the bullet when it leaves the barrel be v.

Thus,

{eq}\begin{align*} K &= \frac{1}{2}mv^{2}\\ \Rightarrow \ v &= \sqrt{\frac{2K}{m}}\\ v &= \sqrt{\frac{2 \times 1091}{0.0300}}\\ v &= 269.69 \ m/s.\\ \end{align*} {/eq}

Here, the bullet starts from the rest, therfore the average speed of the bullet can be given as,

{eq}v_{avg} = \frac{u+ v}{2} {/eq}

{eq}v_{avg} = \frac{0 + 269.69}{2} {/eq}

{eq}v_{avg} = 134.84 \ m/s. {/eq}

Thus, time needed to cross the barrel,

{eq}t = \frac{d}{v_{avg}} {/eq}

{eq}t = \frac{1.00}{134.84} {/eq}

{eq}t = 0.0074 \ s. {/eq}

Thus, the average power delivered to the bullet can be given as,

{eq}P = \frac{K}{t} {/eq}

{eq}P = \frac{1091}{0.0074} {/eq}

{eq}P = 147.43 \ \ k \rm W. {/eq}