Question about work and finding the force of friction

[zstraught]

Homework Statement

A 2.4*10^2 N force is pulling an 85 kg refrigerator across a horizontal surface. the force acts at an angle of 20.0 above the surface. the coefficient of kinetic friction is .200, and the refrigerator moves a distance of 8.00 m. find (a) the work is done by the pulling force, and (b) the work is done by the kinetic frictional force

Relevant Equations

W=Fd

I drew a force diagram, Normal force up, weight down, force of friction to the left, and Force applied 20 degrees above the positive x axis.
I need clarification to find the force of Friction for part B. Should I set Force of friction equal to mu · normal force or should I use Newton's second law and set the force of friction equal to Fa·cos(20)? They both look like valid choices, but they both give different values of force of friction.

 

[haruspex]

set the force of friction equal to Fa·cos(20)?

What about acceleration?

 

[zstraught]

What about acceleration?

Would it be 0 since it is moving at a constant velocity?

 

[kuruman]

Would it be 0 since it is moving at a constant velocity?

Where, in the problem statement, does it say that the velocity is constant? I couldn't find it.

 

[zstraught]

Where, in the problem statement, does it say that the velocity is constant? I couldn't find it.

I thought " A 2.4*10^2 N force..." means it is a constant velocity. If it isn't, how would I find the acceleration?

 

[haruspex]

I thought " A 2.4*10^2 N force..." means it is a constant velocity. If it isn't, how would I find the acceleration?

By finding the net horizontal force and dividing by the mass.
As you wrote:

set Force of friction equal to mu · normal force

 

[Lnewqban]

... Should I set Force of friction equal to mu · normal force or should I use Newton's second law and set the force of friction equal to Fa·cos(20)? They both look like valid choices, but they both give different values of force of friction.

The second choice is not valid.
You could pull that refrigerator with huge force and still couldn't increase the amount of kinetic friction resisting your effort.

That coefficient of kinetic friction being 0.20 means, in practical terms, that once you get that refrigerator moving and continue pulling it across that horizontal surface, the friction force that resists your pulling remains relatively constant at a value of 20% the weight of the refrigerator.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin

 

[kuruman]

$\dots$ the friction force that resists your pulling remains relatively constant at a value of 20% the weight of the refrigerator.

Not quite. The pulling force has a vertical component, therefore the normal force is not equal to the weight.

 

[Lnewqban]

Not quite. The pulling force has a vertical component, therefore the normal force is not equal to the weight.

Exactly!
Thank you, kuruman.
In this case, the pulling-up angle of 20° reduces the normal and friction forces some, as well as the the horizontal component of the pulling force, which is the one doing the positive work in question.