If a #10kg# object moving at #1 m/s# slows down to a halt after moving #3/2 m#, what is the friction coefficient of the surface that the object was moving over?

1 Answer
Feb 13, 2016

Answer:

#mu_s=0.03#

Explanation:

The reason why the object stopped was because the object had to lose its kinetic energy in order to overcome the work done by the surface to stop the ball. This means that what initially was just kinetic energy now is the total energy the surface has dissipated to stop the ball.

i.e #1/2mv_o^2=\mu_sNx#
where #m\implies#mass of object
#v\implies#velocity of object
#N\implies#force of normal
#\mu_s\implies#co-ef of static friction
#x\implies#distance traveled by the object.
We have #m=10kg# #v_o=1# #N=100#(#g# is taken as #10ms^-2#) #x=3/2m#

So, #1/cancel{2}^1*cancel{10}^1*1^2=\mu_s*cancel{100}^10*3/cancel{2}^1#
#implies1=\mu_s*30implies1/30=mu_s#

Solve that small part there and you'll get your anwser.