Box up a ramp - Work and friction force (easy)

[mybrohshi5]

Homework Statement

A 800 N box was pushed up an incline plane of length 4.0m (measured along the incline) to a window ledge 2.0m above ground level. This process required that 3200J of work be done.

What was the average frictional force along the plane?

Homework Equations

W=Fd

The Attempt at a Solution

sin(theta) = 2/4

theta = 30 degrees

W = 800 N (4cos30)

W = 2771 J

W_friction = 3200 - 2771
= 429J

so the force done by friction would just be

429 = F(4cos30)

F = 124 N

This is wrong

the correct answer is 400 N

Thanks for any help :)

 

[diazona]

sin(theta) = 2/4

theta = 30 degrees

W = 800 N (4cos30)

Why did you do this?

After you've thought about that, here's a clue: remember the work-energy theorem.

 

[mybrohshi5]

To be honest i am not sure why i did. i guess i was thinking that Work was force times the displacement of the x component of the ramp.

The work-energy theorem is just W = delta K so W = K_f - K_i which is the same as 1/2mv_f^2 - 1/2mf_i^2

but i don't see how that can help me since i don't know velocity...

 

[diazona]

No, the work-energy theorem is
W = \Delta E
where E is the sum of kinetic and potential energies.

Also, the formula for work in terms of force is
W = \vec{F}\cdot\vec{d}
You multiply the force times the component of displacement parallel to the force. (Or equivalently, multiply the displacement by the component of force parallel to the displacement, which is probably easier in this case)

 

[mybrohshi5]

My book states that the work-energy theorem is just W = delta K ?

I think i am making this problem a lot harder than it actually is but i still can't figure out how to use what you have given me.

I get it will have no potential energy at first but then it will have mgh potential energy when at the top of the ramp.

I am not sure how to approach the kinetic part. i think i need to think that it has kinetic energy at first and then has none at the end cause then i can do

1/2mv^2 = mgh

and solve for the velocity but i don't see how that helps me with finding the friction :(

 

[diazona]

What is the box's speed before it starts moving up the ramp?

What is the box's speed after it has finished moving up the ramp?

By the way, W = \Delta K is called the "work-kinetic energy theorem." I guess some people (and textbooks) do call it the work-energy theorem but then I have to wonder, what name do they use for W = \Delta E, which is arguably the more useful form of the equation?

You could do this problem using W = \Delta K but then you would have to include all forces when computing W, not just friction.

 

[mybrohshi5]

ok so the boxes speed initially can be found using

1/2mv^2 = mgh

v = 6.26 m/s

the boxes final speed is just 0 correct?

i don't feel like i am using the right initial and final kinetic's. This is what always seems to confuse me. I am not sure whether it will have 0 initial and a final kinetic or an initial kinetic and a 0 final or have both an initial and final kinetic.

 

[diazona]

ok so the boxes speed initially can be found using

1/2mv^2 = mgh

v = 6.26 m/s

No. Why would it have that initial speed? Something would have had to accelerate it up to that speed. Besides, that equation does not take into account losses due to friction.

the boxes final speed is just 0 correct?

Yep

 

[mybrohshi5]

So if the boxes speed initially and finally are 0 then

W = U + W_friction

W = mgh

3200 = 800(2) + W_friction

W_friction = 1600 J

but that can't be right cause the answer is 400 J.

I feel like i am being a complete idiot right now :(

 

[diazona]

You're getting warmer!

The answer is not 400J, you said so yourself in the first post.

 

[mybrohshi5]

Oh so if W_friction = 1600J and it is done over a distance of 4 m then

1600 J = F_friction * 4m

F_friction = 400 N

Thank you :)

 

[diazona]

There you go :-)