# Block Projected up Plane with Spring atop

1. Oct 10, 2008

### PatrickR.

1. The problem statement, all variables and given/known data

A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 40 °. A 1.6 kg block is projected up the plane, from an initial position that is distance d = 0.80 m from the end of the relaxed spring. With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by 0.50 m?

2. Relevant equations

Ui + Ki = Uf + Kf

U(elastic)=.5kx2

U(gravitational)= mgy

3. The attempt at a solution

So I assumed that Ui=0 because I'm assuming the block starts at the bottom of the plane.

I also assumed that because the block is momentarily stopping that Kf=0

so that leaves me with

Ki=Uf and solving for Uf will give me Ki

Uf= U(elastic) + U(gravitational)

Uf = .5(200N/m)(.5m)2 + (1.6kg)(9.8m/s2)([.5m+.8m] sin 40)

Uf= 38.1026J

therefore Ki should equal 38.1026

However, that answer is refused so I would like to know what I am doing wrong. Thankyou.

2. Oct 10, 2008

### Rake-MC

You have a good start. Do you know why you took the sin(40) in the fourth last line?

This is because the gravitational potential is $$mg\Delta h$$

By taking this sine, you are making the change in height parallel to the force such that you get the right answer. However you also took the sine for spring potential energy. However the line of force of the spring is parallel to the plane.

3. Oct 10, 2008

### PatrickR.

Even more troubling, I still got the answer wrong and it was my last attempt so they just show you how to do it then and my setup actually was correct however the computer did not want to take my numbers (the numbers are individualized for everyone.

Rake I do actually think the right way to do it is what you said and not consider the .5 into the sin 40
so, it should look like this

Uf = .5(200N/m)(.5m)2 + (1.6kg)(9.8m/s2)([.5m]+.8 sin 40)

but this just makes me hate physics even more...