Finding work and the Change in Thermal Energy

[alyssad]

Homework Statement

Kerry is pulling a 156-kg sled along a snowy, horizontal path with a 620-N force directed at an angle of 29.0° above the ground. He pulls the sled over a distance of 26.0 m, and the coefficient of kinetic friction between the sled and the ground is 0.0622.

I need to find the amount of work that Kerry does as he pulls the sled AND I need to find the increase in thermal energy.

Homework Equations

I know that
W = μ_kmgΔr*cos(Θ)
ΔE_th = F_kΔr

The Attempt at a Solution

I'm not really sure where to begin with finding the answer. I feel like the work equation needs to include the force that is given in the problem statement. And I would appreciate help with the thermal energy part as well.

Thank you!

 

[Chestermiller]

Homework Statement

Kerry is pulling a 156-kg sled along a snowy, horizontal path with a 620-N force directed at an angle of 29.0° above the ground. He pulls the sled over a distance of 26.0 m, and the coefficient of kinetic friction between the sled and the ground is 0.0622.

I need to find the amount of work that Kerry does as he pulls the sled AND I need to find the increase in thermal energy.

Homework Equations

I know that
W = μ_kmgΔr*cos(Θ)
ΔE_th = F_kΔr

The Attempt at a Solution

I'm not really sure where to begin with finding the answer. I feel like the work equation needs to include the force that is given in the problem statement. And I would appreciate help with the thermal energy part as well.

Thank you!

Have you drawn a free body diagram?

 

[alyssad]

Have you drawn a free body diagram?

I have drawn a free body diagram. I'm pretty sure that the equation I posted is the equation for the amount of work needed to overcome the force of friction. I still don't know where to go from here.

 

[Chestermiller]

Is the sled accelerating, such that its kinetic energy is increasing?

 

[alyssad]

So I found that W = 14098 J, and I know that is the correct answer. How should I start finding the change in thermal energy?

 

[Chestermiller]

The change in thermal energy is equal to the frictional work.