On the horizontal surface there is a block of m = 0.9 kg. it gets hit by a bullet m = 12g flying horizontally

On the horizontal surface there is a block of m = 0.9 kg. it gets hit by a bullet m = 12g flying horizontally V = 300 m / s and gets stuck in it. Find the coefficient of sliding friction force if the bar passes the path L = 11m before it stops completely

Given:

m1 = 0.9 kilograms – the mass of the bar;

m2 = 12 grams = 0.012 kilograms – the mass of the flying bullet;

v = 300 m / s – the speed of the flying bullet;

L = 11 meters – the path traversed by the bar after being hit by a bullet;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine the coefficient of friction k.

Let’s find the speed of the bar after the bullet has hit it:

m2 * v = (m1 + m2) * v2;

v2 = m2 * V / (m1 + m2) = 0.012 * 300 / (0.9 + 0.012) = 3.6 / 0.912 = 3.9 m / s.

Then the acceleration with which the bar braked is equal to:

a = v2 ^ 2 / (2 * L) = 3.9 ^ 2 / (2 * 11) = 15.21 / 22 = 0.7 m / s ^ 2.

This acceleration was imparted to the bar by a frictional force, which is equal to:

F = (m1 + m2) * a = 0.912 * 0.7 = 0.64 Newtons.

Then the coefficient of friction is:

k = F / ((m1 + m2) * g) = 0.64 / (0.912 * 10) = 0.64 / 9.12 = 0.07.

Answer: The coefficient of friction is 0.07.