# Work Energy Theorem+Projectile motion

1. Oct 31, 2007

### rwx1606

1. The problem statement, all variables and given/known data
A 25.0 g mass rests on a 50.0 inclined plane (uk = 0.200). It is pushed against a spring, compressing it 15.0 cm, and released. It moves up the plane and leaves the spring, traveling an additional 45.0 cm before leaving the edge of the plane (at a height of 80.0 cm above the ground). It then flies through the air, and is observed to land 270. cm from the base of the incline. What was the spring constant?

2. Relevant equations
Wnet = ΔK, mvf^2 - mvi^2
Xf = Xi + Vit + (1 / 2)at2

3. The attempt at a solution
Well, what I thought was the key to solving this problem was to realize the final speed on the incline is the initial speed of a projectile motion problem. So I found the work done by the spring, gravity,and friction. The net work I calculated was 0.1125k-13.2 (where k is the spring constant). I set this equation equal to the change in kinetic energy. I know the object had an initial velocity zero, and as mentioned before, the final velocity should be equal to the initial velocity of the projectile motion part of the problem. So I solved for the final velocity at the top of the incline by using the kinematics equation. I assumed mass was launched at the same angle as the incline, 50.0 degrees. After getting the final initial velocity from this projectile part of the problem, I set that equal to the final velocity of the mass at the top of the incline. It turned out to be 0.1125k-13.2=1/2(.0250kg)(21.5m/s) and found the spring constant to be 1192.96 N/m. I only received half credit for the problem and was hoping you guys could help me figure out where I went wrong. I checked my math a couple of times, so I think I'm making a conceptual mistake. Any help is appreciated!

2. Oct 31, 2007

### Staff: Mentor

That's good.
This is the work done on the object as the spring decompresses? Or for the entire slide up the incline? Show how you arrived at this number. (Give the equations and what you used for the variables.)

3. Oct 31, 2007

### rwx1606

The equation I used for the work done BY the spring is 1/2k(Xi^2 - Xf^2). The initial position I put -.15m and the final position 0. So, 1/2k(-.15^2 - 0) is the work done by the spring, which i got .01125 k. For the work done by gravity I used W=Fs cos(theta). Which turned out to be (9.8)*(.025)*cos(90+50)*cos(180)*60. So the work done by gravity is -11.3 J. The cos(90+50) is the angle between the displacement and the force. AH! as I'm typing this I realized I put 60m as the distance of the strip and forgot to convert to cm. I'm going to recalculate. Can someone get an answer that I can confirm with?

4. Oct 31, 2007

### rwx1606

I recalculated and got 35.6 N/m for the spring constant.