# Homework Help: Unknown mass being pushed up ramp until it hits spring

1. Feb 23, 2012

### maxpower612

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

A force F pushes a block of mass m up a frictionless incline as shown below. The block moves
at a constant speed of 6.00 m/s until it contacts the relaxed spring which has a spring constant k =
200 N/m. At the point where the block touches the spring, the force is removed. The spring is then compressed by 0.60 meters until the block comes to rest.
What is the mass of the object in kg?
It is at an angle of 40 degrees. g=9.8 m/s2
The height and length are not given but I'm assuming Hi=0

Also in what direction would the net force be on the block when the block is at rest and fully compressed?
2. Relevant equations
Us=1/2kx2 (1)
Ki+Usi=Kf+Usf (2)

3. The attempt at a solution

I inserted the values into equation 2 and got 2 kg but I don't think it's right since on an incline, there must be potential energy involved.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 23, 2012

### PeterO

The last question is really easy.

In what direction will the mass move?

It it is about to bury itself vertically into the surface, then the net force is down.
If it is about to spring vertically up, then the net force is up.
If it is about to fly off the slope horizontally, then the net force is horizontal.
If it is about to move down the slope, then the net force is down the slope, parallel to the slope.
And if it is going to stay where it is - then the net force is zero.

3. Feb 23, 2012

### PeterO

In red above ... True.

Overview:

When the mass reaches the spring, force F ceases, so we don't have to worry about that.

The mass was moving, so had kinetic energy.

When it finally stops, it will be at a higher elevation - so has gained some PE, and has compressed the spring and stored some energy in there [another gain].
The two gains must total the loss of Kinetic energy as it stops..

4. Feb 23, 2012

### maxpower612

I wasn't sure in what direction the net force was since the object was at rest, but the net force moving parallel to the slope seems the most logical answer. Thanks for your reply!

5. Feb 23, 2012

### maxpower612

So would the equation be 1/2mvi^2+0+0=0+mgHf+1/2kx^2? if so how would I find the final height in order to get the mass?

6. Feb 23, 2012

### PeterO

The height is a component of x.

7. Feb 23, 2012

### maxpower612

I took your advice (assuming I did it right) and found the height to be (0.6m)sin40= .386m and plugged into back into the previous equation by getting the mass approximately equal to 2.53 kg. I hope that's right lol