# If a 2 kg object moving at 5 m/s slows down to a halt after moving 100 m, what is the friction coefficient of the surface that the object was moving over?

Jan 24, 2018

$0.0125$

#### Explanation:

Here,if constant retardation occurring due to frictional force is a then we can use the equation ${v}^{2} = {u}^{2} - 2 a s$(all symbols are bearing their conventional meaning) where, $v = 0$ ,$u = 5 \frac{m}{s}$ and $s = 100$

So, $a = 0.125 \frac{m}{s} ^ 2$

So,here amount of force acting against the motion of the object is m×a or (2×0.125) N

And this will be equal to the amount of frictional force acting at their interface i.e mu×N = mu×mg (where, $\mu$= coefficient of frictional force)

So,we can write,

mu×2×10 = 2 × 0.125 or $\mu$ = $0.0125$