# Homework Help: Conservation of Energy equation help

1. Jan 4, 2006

### PrudensOptimus

Code (Text):

A 12kg block is released from rest on a 30degree frictionless incline. Below the block is a spring that can be compressed 2.0cm by a force of 270N. The block momentarily stops when it compresses the spring by 5.5 cm.

(a). How far does the block move down the incline from its rest position to this stopping point?

(b). What is the speed of the block just as it touches the spring?

k = -F/x = -mg / x = 135N/cm

Since Wy = 0

Wx = W = Wg + Wspring = mgsin30dx - kdx^2/2

Then... i am lost loll...

2. Jan 4, 2006

### dicerandom

You're on the right track! I'm not quite sure what your dx's are though, are those differentials or distances?

I think of it this way: Initially the block has entirely potential energy, mgh. When the block comes to rest all of its energy is now stored in the potential of the spring, 1/2 k x^2. Since the plane is frictionless energy is conserved and therefore the initial and final energies must be equal.

You calculated k, x and m were given in the problem, and g's just a constant so you can find h and then use the angle given to find the distance which the block slid on the plane.

3. Jan 4, 2006

### PrudensOptimus

The force in the spring equation... is it Force of gravity in the x direction only in this case??? ( since the sum of all forces is only the Force of gravity in the x direction )

4. Jan 4, 2006

### PrudensOptimus

so it is

mgh = kx^2/2 ????

solve for x ?

5. Jan 4, 2006

### Astronuc

Staff Emeritus
From "Below the block is a spring that can be compressed 2.0cm by a force of 270N" one can compute the spring constant k.

Edit: Refer to dicerandom's post below.

The block is at some height, so has some gravitational potential energy. When it is released, the block accelerates under gravity, and some of the gravitational potetial energy in converted to kinetic energy. Then the block strikes the spring and immediately, the spring force begins to increase, and the kinetic energy is converted into the mechanical potential energy of the spring AND some additional gravitational potential energy (because the block is still decreasing in elevation) is converted into spring mechanical energy.

But with conservation of energy (assuming no friction of block on incline and not internal friction in the spring), one the block stops, then the entire gravitational potential energy must be converted into spring stored energy, and mgh = kx2/2 is correct where x is the displacement from the reference point (i.e. no load).

Last edited: Jan 4, 2006
6. Jan 4, 2006

### dicerandom

x is the displacement of the spring, they gave you that in the problem. You need to solve for h and then use that to find the distance that the block slid down the incline.