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

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\impliesmass of object
v\impliesvelocity of object
N\impliesforce of normal
\mu_s\impliesco-ef of static friction
x\impliesdistance 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.