# 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)mv^{2}

F_{F} = u_{k} * F_{N}

W = f * d

v^{2} = vi^{2} + 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)mv^{2}

KE = (1/2) * (30) * 4^{2}

KE = 240J

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

KE = (1/2)mv^{2}

KE = (1/2) * (30) * 2^{2}

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:

v^{2} = vi^{2} + 2a(x – xi)

4 = 0 + 2*a*3

a= (2/3)

Now I sum the forces:

20N – (u_{k} * (30kg * 9.8)) = 30kg * (2/3)

u_{k} = 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.

Could anyone please help me understand what’s incorrect?

Thanks.

http://ift.tt/1d5TFXK

## Leave a comment