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FOR THE GIVEN PROBLEM AND SOLUTION BELOW

1) The problem statement, all variables and given/known data: A student pulls on a box up an incline of 10 degrees. The student pulls with a force of 180 N directed at 30 degrees above the line of the incline. The box has a mass of 28 kg, and the coefficient of friction between the box and floor is 0.299. The acceleration of gravity is 9.8 m/s^2.

What is the acceleration of the box?

2) Relevant equations Fnet = ma

3) The attempt at a solution From Fnet = ma I get: Fp - Ffr = ma, where Fp is the force of the person pulling and Ffr is the force of friction resisting.

Ffr = MkFn Ffr = (.299)(Fn)

I then solved for Fn: Fn = mg/cos10 Fn = (28)(9.8) / cos10 Fn = 278.63

So Ffr = (.299)(278.63) Ffr = 83.31 N

So we have Fp - Ffr = ma, so Fp - 83.31 = (28)a

We need Fp. Fp = cos30 * 180 Fp = 155.88

So Fp - 83.31 = (28)a, so 155.88 - 83.31 = (28)a

Solving for a i get a = 2.59 m/s^2

QUESTION If the value of Fp in this problem is not stated, how can I compute for the work done by Fp?

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