How to Calculate Work Done by Kinetic Friction on a Toboggan?

[Jimmer7]

Homework Statement

A toboggan carrying two children with a total mass of 85.0 kg reaches its maximum speed at the bottom of a hill, and then glides to a stop in 21.0 m along a horizontal surface. The coefficient of kinetic friction between the toboggan and the snow surface is 0.110. Calculate the work done by the kinetic friction?

Homework Equations

W=FCosθΔd

The Attempt at a Solution

m = 85 kg
Δd = 21.0 m
θ = 0
W = ?
Mu = 0.110

I'm pretty confused on this, do I somehow get Mu to Ff and then plug it into the work equation?

The final answer is -1.90 x10^3 J but from what I've tried I never got it.

 

[gneill]

Homework Statement

A toboggan carrying two children with a total mass of 85.0 kg reaches its maximum speed at the bottom of a hill, and then glides to a stop in 21.0 m along a horizontal surface. The coefficient of kinetic friction between the toboggan and the snow surface is 0.110. Calculate the work done by the kinetic friction?

Homework Equations

W=FCosθΔd

The Attempt at a Solution

m = 85 kg
Δd = 21.0 m
θ = 0
W = ?
Mu = 0.110

I'm pretty confused on this, do I somehow get Mu to Ff and then plug it into the work equation?

The final answer is -1.90 x10^3 J but from what I've tried I never got it.

There's no angle in the problem, so your Relevant Equation is not what you need here.

The toboggan is traveling horizontally as it slows to a stop due to friction. So, it looks like you'll need to find the frictional force acting (what equation will you apply?) and then an equation relating force and distance to work done. What equation might that be?

 

[Jimmer7]

There's no angle in the problem, so your Relevant Equation is not what you need here.

The toboggan is traveling horizontally as it slows to a stop due to friction. So, it looks like you'll need to find the frictional force acting (what equation will you apply?) and then an equation relating force and distance to work done. What equation might that be?

I tried it and I got it right. :)

I did:
FF = M.FN
FN = mg = 85 (9.8)
FF = 0.110 (85 x 9.8) = 91.7235 N
W = (91.7235)(0)(21) = 1926.1935
W = -1.90 x 10^3 J

made it cos180 which made it negative.

Thanks.