Question

In a ballistics test, a 28-g bullet pierces a sand bag that is 30 cm thick. If the initial bullet velocity was 55 m/s and it emerged from the sandbag moving at 18 m/s what was the magnitude of the friction force (assuming it to be constant) that the bullet experienced while it traveled through the bag?

Answer 1

Answer:

Frictional force will be equal to 126.04 N

Explanation:

We have given mass of the bullet m = 28 gram = 0.028 kg

Initial velocity u = 55 m /sec

Width of sand = 30 cm = 0.3 m

So initial kinetic energy $E=\frac{1}{2}mu^2=\frac{1}{2}\times 0.028\times 55^2=42.35j$

Final velocity of the bullet v = 18 m /sec

So final kinetic energy $E=\frac{1}{2}mv^2=\frac{1}{2}\times 0.028\times 18^2=4.536j$

So change in kinetic energy = 42.35 - 4.536 = 37.814 j

From work energy theorem this change in kinetic energy will be equal to work one by frictional force

So $f\times 0.3=37.814$

$f=126.04 N$