# Inclined plane problem

1. Feb 10, 2008

### NAkid

[SOLVED] inclined plane problem

1. The problem statement, all variables and given/known data
A 3.4kg bundle starts up a 30o incline with 198J of kinetic energy. How far will it slide up the plane if the coefficient of friction is 0.26?

2. Relevant equations
Work(total)=.5mv(final)^2 - .5mv(initial)^2

3. The attempt at a solution
I set up a FBD for work(friction and work(gravity) to calculate work(total)=k2=k1
work(friction)=-ukmgsin(theta)=-4.3s (s=displacement)
work(gravity)=-mgs=-33.32s (s=displacement)
work(total)=-37.65s=k2-k1
k2=.5mv(final)^2=0 because vf=0
k1=198J
so s=198/37.65 --- but that's not the right answer.

2. Feb 10, 2008

### mgb_phys

Energy is force * distance.
What is the frictional force parralel to the surface the block slides on?

Then account for the potential energy added to the block.

3. Feb 10, 2008

### NAkid

well, frictional force is -ukmgsin(theta) right?

4. Feb 10, 2008

### NAkid

i guess my question is, what is wrong with my logic?

5. Feb 10, 2008

### mgb_phys

Have you drawn a force diagram? Are you sure about the sin()?
Rather than try and memorise formula think what happens - if the incline is almost level is the friction highest or lowest, does the frictional force get larger or smaller as the incline goes to vertical?

6. Feb 10, 2008

### NAkid

i guess that depends on whether it's a sin or a cos. if it's sin the frictional force should get higher as the incline goes to vertical, because the sin(theta)-->1 towards the vertical, but lower if cos because cos(theta)--> 0 towards the vertical.

7. Feb 10, 2008

### NAkid

and with that, it's supposed to be -ukmgcos(theta)

8. Feb 10, 2008

### NAkid

which still doesn't yield the right answer

9. Feb 10, 2008

### mgb_phys

The angle of an inclined plane is normally measured from the horizontal - which by similair triangles is also the angle between the normal force on the object and vertical.
Since the normal force is the full weight when the plane is horizontal then the term must be cos() since sin(0)=0.

So you have a resistance to movement of ukmgcos(theta) together with a distance along the plane which gives you energy.
There is also a vertical distance = length*sin(theta) so you have a potential energy gain.

10. Feb 10, 2008

### NAkid

ah of course - my W(grav) set up was wrong, it should be -mgsin(theta)*s

thank you very much!!