# If an object with a mass of 10 kg  is moving on a surface at 45 m/s and slows to a halt after  4 s, what is the friction coefficient of the surface?

Jun 6, 2018

The coefficient of friction is $= 1.15$

#### Explanation:

Calculate the acceleration with the equation

$v = u + a t$

The initial velocity is $u = 45 m {s}^{-} 1$

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

The time is $t = 4 s$

The acceleration is

$a = \frac{v - u}{t} = \frac{0 - 45}{4} = - 11.25 m {s}^{-} 2$

The mass of the object is $m = 10 k g$

Apply Newton's Second Law to calculate the frictional force

${F}_{r} = m \cdot | a | = 10 \cdot 11.25 = 112.5 N$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

The normal force (reaction) is $N = m g = 10 \cdot 9.8 = 98 N$

The coefficient of friction is

${\mu}_{k} = {F}_{r} / N = \frac{112.5}{98} = 1.15$