# If an object is moving at 120 m/s over a surface with a kinetic friction coefficient of u_k=12 /g, how much time will it take for the object to stop moving?

Oct 9, 2017

The object will stop in 10 seconds. Complete solution follows...

#### Explanation:

We must start by determining the acceleration of the body, which we do by considering the force picture. Since the only horizontal force is friction, Newton's second law becomes

${F}_{\text{net}} = {F}_{f} = m a$

where ${F}_{f} = \mu {F}_{N} = \mu m g$

(if the motion is on a horizontal surface, it will be true that ${F}_{N} = m g$)

Combining these two equations, we get

$m a = \mu m g$ or simply $a = \mu g$

and, since we are told that $\mu = \frac{12}{g}$, we can express the acceleration as

$a = 12 \frac{m}{s} ^ 2$ (actually $- 12 \frac{m}{s} ^ 2$ as teh object will be slowing down)

Now use the equation of motion ${v}_{f} = {v}_{i} + a \Delta t$

Since ${v}_{f} = 0 , {v}_{i} = 120 \mathmr{and} a = - 12 ,$ we get

$0 = 120 - 12 \Delta t$

so that $\Delta t$ must be 10 seconds.