Question

A 0.020-kg bullet traveling at a speed of 300 m/s embeds in a 1.0-kg wooden block resting on a horizontal surface. The block slides horizontally 4.0 m on a surface before stopping. What is the coefficient of friction between the block and the surface?

Answer 1

Answer:

The coefficient of friction between the block and the surface is μk=0.441

Explanation:

Let

mb = bullet mass = 0.02 kg

vb = bullet speed  = 300 m/s

mw = mass of the wood block = 1 kg

d = distance covered by the block plus the bullet = 4 m

u = speed of the bullet plus the block

g = gravity = 9.8 m/s2

μk = coefficient of friction, variable unknown

We will use the following expression to calculate the coefficient of friction:

μk = $\frac{(\frac{mb*vb}{mb+mw})^{2} }{2*g*d}$

Replacing known values:

μk=$\frac{(\frac{(0.02kg)*(300m/s)}{0.02kg+1kg})^{2} }{2*9.8m/s^{2}*4m }=0.441$

The coefficient of friction is a dimensionless value