# If an object with a mass of 10 kg  is moving on a surface at 15 m/s and slows to a halt after  4 s, what is the friction coefficient of the surface?

Apr 27, 2017

The coefficient of friction is $= 0.38$

#### Explanation:

We start by calculating the deceleration.

We apply the equation of motion

$v = u + a t$

The initial velocity is $u = 15 m {s}^{-} 1$

The time is $t = 4 s$

The final velocity is $v = 0 m {s}^{-} 1$

Therefore,

$0 = 15 + 4 a$

The acceleration is

$a = - \frac{15}{4} m {s}^{-} 2$

According to Newton's Second Law, the frictional force is ${F}_{r} = m a$

So,

${F}_{r} = 10 \cdot \frac{15}{4} = \frac{75}{2} N$

The normal force is

$N = m g = 10 \cdot 9.8 = 98 N$

The coefficient of friction is

${\mu}_{k} = {F}_{r} / N = \frac{37.5}{98} = 0.38$