# Spring Compression and block of mass

1. Oct 7, 2008

### Husker70

1. The problem statement, all variables and given/known data
A block of mass 12.0kg slides from rest down a frictionless 35degree
incline and is stopped by a strong spring with a force constant of
3.00 x 10^4 N/m. The block slides 3.00m from the point of release
to the point where it comes to rest against the spring. When the block
comes to rest, how far has the spring been compressed?

2. Relevant equations
W=deltaK
Vf^2 = vi^2 + 2a(deltay)
Wg + Ws = delta K
-mg + 1/2kx^2 = 1/2m(vi^2)

3. The attempt at a solution

I drew my picture and forces and then
I solved for the velocity at the moment it makes contact at the spring
Vf = sqrt(Vi^2 + 2(9.8m/s^2)(sin35)(3.0m) = 5.81m/s

Then I used -mg + 1/2kx^2 = 1/2m(vi^2)
I rearranged to be
x = sqrt 2(1/2mv^2 + mg)/K

x = sqrt 2(1/2)(12kg)(5.81)^2 + (12kg)(9.8m/s^2)(sin35) / 3.00 x 10^4 N/m

x = 540/30000
x = .018m
The book says .116m
I must not be on the right track. Any help would be appreciated.
Kevin
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 7, 2008

### Rake-MC

What you have wrong here is you are mixing units. You have force - energy = energy.

Here is a simple way to look at the problem with conservation of energy.

When the object is released, all energy is gravitational potential energy.
When it comes to rest, on the spring, all energy has become spring potential energy (if you take the maximum compression of the spring to he h=0).

Therefore write your equation as: $$mg\Delta h = \frac{1}{2} kx^2$$

You know m, g, h and k. Solve for x.

3. Oct 7, 2008

### Husker70

That makes more sense. I'll try that.
Thank you,
Kevin

4. Oct 7, 2008

### Husker70

Is the h sin35?
Kevin

5. Oct 7, 2008

### Rake-MC

not just sin(35), it's 3sin(35) because it's the height equivelant of 3 metres down the plane.

6. Oct 7, 2008

### Husker70

That makes sense.
Thanks again,
Kevin