# Coefficient of Friction and Energy

1. The problem statement, all variables and given/known data

A 30kg sled is sliding on a frictionless sheet of ice at a velocity of 4m/s. The sled encounters a rough patch of ice and begins to slow down. After traveling on the rough patch of ice for 3m, the sled’s velocity is 2m/s. Determine the coefficient of friction between the rough ice and sled.

2. Relevant equations

KE = (1/2)mv2

FF = uk * FN

W = f * d

v2 = vi2 + 2a(x – xi)

3. The attempt at a solution

I begin by finding the Kinetic energy when the sled is on the frictionless sheet of ice:

KE = (1/2)mv2

KE = (1/2) * (30) * 42

KE = 240J

Then I find the Kinetic Energy when the sled is on the rough patch:

KE = (1/2)mv2

KE = (1/2) * (30) * 22

KE = 60J

I now note the amount of energy released due to friction:

240J – 60J = 180J

I continue to evaluate the force acting on the sled when on the rough patch:

W = fd
60J = f * 3m
f = 20N

Then I solve for acceleration in the rough patch:

v2 = vi2 + 2a(x – xi)

4 = 0 + 2*a*3
a= (2/3)

Now I sum the forces:

20N – (uk * (30kg * 9.8)) = 30kg * (2/3)
uk = 0

This is obviously not correct as it states friction is present. I don’t know where I’m making a mistake. Presumably my logic is incorrect, not the math itself. I think finding the kinetic energy for the frictionless surface was useless with the approach I’m taking.