If an object is moving at #120 m/s# over a surface with a kinetic friction coefficient of #u_k=7 /g#, how much time will it take for the object to stop moving?

1 Answer
Dec 21, 2015

Answer:

I found #17.1sec#

Explanation:

Kinetic friction #f_k# is responsible for introducing a deceleration given by (Newton's Second law):
#-f_k=ma#
#-u_k*N=ma#
#-u_k*cancel(m)g=cancel(m)a#
So:
#a=-7/g*g=-7#
We can now use:
#color(red)(v_f=v_i+at)#
To get:
#0=120-7t#
#t=120/7=17.1 sec#