A bullet of mass $15\ g$ has a speed of $400\ m/s$. What is its kinetic energy? If the bullet strikes a thick target and is brought to rest in $2\ cm$, calculate the average net force acting on the bullet. What happens to the kinetic energy originally in the bullet?

Here given, mass of the bullet $=15\ g=\frac{15}{1000}\ kg$

Speed of the bullet $v=400\ m/s$

Therefore, kinetic energy $K=\frac{1}{2}mv^2$

$=\frac{1}{2}\times\frac{15}{1000}\times(400)^2$

$=1200\ J$

So, the kinetic energy of the bullet is $1200\ J$.

Now, Let $F$ be the net force acting on the bullet.

So, work done by the net force in $2\ cm$ will be equal to the change in kinetic energy.

Work done$=$Change in kinetic energy

Or $F\times d=K_i-K_f$     [$K_f\rightarrow$Final kinetic energy and $K_i\rightarrow$initial kinetic energy]

Or $F\times \frac{2}{100}\ m$=1200\ J-0=1200\ J$   [As $K_f=0$ because the bullet came to rest.]

Or $2F=120000$

Or $F=\frac{120000}{2}$

Or $F=60000\ N$

Or $F=6.0\times10^4\ N$

Therefore, $6.0\times10^4\ N$ net force acted on the bullet in $2\ cm$.

The kinetic energy of the bullet is converted into heat energy.

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Updated on: 10-Oct-2022

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