Question

A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coefficient of kinetic friction between the stone and the surface?

Answer 1

Answer:

Coefficient of friction will be 0.296

Explanation:

We have given initial speed of the stone u = 8 m /sec

It comes to rest so final speed v = 0 m /sec

Distance traveled before coming to rest s = 11 m

According to third equation of motion

$v^2=u^2+2as$

So $0^2=8^2+2\times a\times 11$

$a=\frac{-64}{22}=-2.90m/sec^2$

Acceleration due to gravity $g=9.8m/sec^2$

We know that acceleration is given by

$a=\mu g$

So $2.90=9.8\times \mu$

$\mu =\frac{2.9}{9.8}=0.296$

So coefficient of friction will be 0.296