Work done by kinetic friction force without coefficient?

[Jessica Sweet]

Homework Statement

A person pulls a box (m=10 kg) horizontally with +2m/s² acceleration by applying 45 N force. The displacement of the box is 8 meters from initial position.
a) How much work is done by applied force?
b) How much work is done by kinetic frictional force?

Homework Equations

Wf = Fcosθs
F=ma
Wfk = -Fks ?
Fk = μN ?

The Attempt at a Solution

I solved for part a, which was 360J.
I don't know how to solve for part b, which it seems like I would need the friction coefficient?
I tried Wfk = -(10)(2)(8) = -160 J, but that is incorrect.
How do I find Fk to solve for the work done by friction?

 

[jack action]

Answer this question: How much of the applied force is used to accelerate the box?

 

[Jessica Sweet]

Um, 45 N of force was applied to it. Would force of kinetic friction be -45 N then? Or am I just getting myself confused?

 

[Entanglement]

Jack has a point.
Yes, you applied 45 N,
but apparently that's not the force used to accelerate the box since its mass is 10 Kg and its accelerations is 2 m / s / s. The net force is what you should be looking at. Rethink. :):)

 

[jack action]

\sum F = ma
F_{total} - F_{friction} = ma

 

[nasu]

Write Newton's second law for the box.

 

[Jessica Sweet]

Alright, so F_total = 45 N, I am looking for frictional force, and ma = 20.
45-F_fk = 20
Ffk=25
Then I have Wfk = -(25)(8) = -200 J?
Is this the correct answer?

 

[jack action]

Look again at your OP and you already found 360 J and 160 J. How does your 200 J fits in there?

Just for the fun of it, can you find the coefficient of friction?

 

[Jessica Sweet]

Are you saying I could have subtracted the 160 from 360? What value does the 160 J represent that I solved for? Gosh, I feel stupid.

 

[Jessica Sweet]

Never mind, I got it all covered. My brain is working again. It's amazing what stress can do to the brain. Thanks for the help everybody.