- #1

Blakeasd

- 11

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## Homework Statement

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.

## Homework Equations

KE = (1/2)mv

^{2}

F

_{F}= u

_{k}* F

_{N}

W = f * d

v

^{2}= vi

^{2}+ 2a(x - xi)

## 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.

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