# If a 3kg object moving at 15m/s slows down to a halt after moving 25 m, what is the friction coefficient of the surface that the object was moving over?

Jan 14, 2017

The friction coefficient must be 0.46 to cause this motion.

#### Explanation:

First, we must find the acceleration of the body:

${v}_{f}^{2} = {v}_{i}^{2} + 2 a \Delta d$

$0 = {15}^{2} + 2 a \left(25\right)$

$a = - \frac{225}{50} = - 4.5 \frac{m}{s} ^ 2$

Now, we can calculate the net force acting on the object from Newton's Second Law

${F}_{\text{net}} = m a = \left(3 k g\right) \left(- 4.5 \frac{m}{s} ^ 2\right) = - 13.5 N$

(The negative sign means that the force opposes the direction of motion, so we will disregard it from here on.)

Since friction is the only force acting, it is also equal to the net force. Therefore

$\mu m g = 13.5 N$

$\mu$ = 13.5/((3)(9.8)) = 0.46