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?

1 Answer
Oct 9, 2017

Answer:

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_"net" = F_f=ma#

where #F_f = muF_N = mu mg#

(if the motion is on a horizontal surface, it will be true that #F_N = mg#)

Combining these two equations, we get

#ma=mumg# or simply #a=mug#

and, since we are told that #mu=12/g#, we can express the acceleration as

#a=12m/s^2# (actually #-12 m/s^2# as teh object will be slowing down)

Now use the equation of motion #v_f = v_i +aDeltat#

Since #v_f=0, v_i=120 and a = -12, # we get

#0=120-12Deltat#

so that #Deltat# must be 10 seconds.