# Box moving up a slope by a force

1. Nov 17, 2012

### BWE38

A 40.8 kg box is being pulled up a 13.5 deg slope by a force of 294 N which is parallel to the slope. The coeffcient of kinetic friction between the box and the slope is 0.21. what is the acceleration of the box?
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Equations: F=MA, Fk=μk(MA)
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Attemped at the solution

deg= 13.5 M= 40.8 kg F=294 μκ= 0.21

force x-direct---------------------------------------force y-direct

294(cos13.5)(0.21)=(40.8 kg)A------------------294(sin13.5)(.21)=(40.8)A
60.03=40.8A---------------------------------------14.41=40.8A
A=1.47 m/s^2-------------------------------------A= 0.35 m/s^2

acceleration= √(1.47^2+0.35^2)
acceleration= 1.51 m/s^2
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The answer is 2.91 m/s^2, which I don't understand.

Last edited: Nov 17, 2012
2. Nov 17, 2012

### ap123

I don't really understand your calculations here.

Have you drawn a free-body diagram with all 4 forces? Put the x-axis along the slope of the incline, and then show us what your expressions for the x and y components are.

3. Nov 17, 2012

### BWE38

I have my free body diagram attached.

#### Attached Files:

• ###### free body diagram.png
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4. Nov 17, 2012

### ap123

For the y-direction ( perpendicular to slope ),
normal force = weight * cos(13.5) = 389N

For the x-direction ( up the slope ),
applied force - friction - weight_x = m*a

What do you get when you work this out?
( remember that friction = μ * normal force )

5. Nov 17, 2012

### BWE38

normal force=40.8cos13.5(9.81)=389.19N

friction=μκ(Nf)=81.73 N

x-direction= applied force - friction - weight= MA

294-81.73-40.8=40.8A
212.27 N - 40.8 kg= 40.8 kg A

am I doing everything right so far?

6. Nov 17, 2012

### ap123

This is all good apart from the weight part - you need the x-component of the weight, ie mgsinθ.
Once you put that in, it will work out correctly.