posted by Lindsay on .
A hockey puck is sliding across a frozen pond with an initial speed of 6.5 m/s. It comes to rest after sliding a distance of 7.6 m. What is the coefficient of kinetic friction between the puck and the ice?
I know this one is not that difficult. However, I can't seem to get the correct answer. Any help is much apprectiated.
Initial puck kinetic energy
= work done against friction
(1/2) M V^2 = M*g*muk*X
muk is the coefficient of kinetic friction that you want. M cancels out, so you don't need to know it.
X is the distance that the puck slides while coming to a stop.
muk = (1/2) V^2/(g X)
That's all there is to it!
Ahh ok, I didn't realize the masses cancel out. But of course you're right.
That really was easy. :)
given:mass m= 20 kilogram ,force = 50 N from newton 2nd law,
f=ma where a=acceleration of the box
a=f over m= 50 over 20=2.5