# Speed, distance and spring on a track

1. Jun 7, 2009

### brunettegurl

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

Tarzan (m=90kg) presses against a horizontal spring with spring constant k=81000N/m ( on a frictionless surface), compressing it by 100 cm. After loosing contact with the spring, Tarzan starts going down an 11.0 m high hill that is frictionless, and then starts to go up a frictionless incline of 30o on the other side. How far along the slope will Tarzan get?
Picture can be found here https://www.physicsforums.com/showthread.php?t=315325"

2. Relevant equations

e=0.5mv2
eg=mgh
spring = 0.5kx2

3. The attempt at a solution

this question very similar to a previous question i posted but i can not seem to carry it forward and find the distance the person would travel it they went up the hill of the frictionless hill. i first found the speed which was 30 m/s. since the track is frictionless i assummed the speed would be the same and set up and another equation 0.5mv2=mgh where h= dsin$$\vartheta$$ and i solved for d this didnt work out ..pls. help

Last edited by a moderator: Apr 24, 2017
2. Jun 7, 2009

### Noein

The speed wouldn't be the same all along the track. Tarzan accelerates down the hill and decelerates up the ramp.

3. Jun 7, 2009

### brunettegurl

so do i find the acceleration of the person using a kinematic equation where the acclerration is gravity - x. x being the deceleration>>??

4. Jun 7, 2009

### dx

Use conservation of energy. Acceleration is not needed here.

5. Jun 8, 2009

### brunettegurl

if i used conservation of energy then wouldn't this be right 0.5mv2=mgh where h= dsin$$\vartheta$$

6. Jun 8, 2009

### dx

You forgot the initial gravitational potential energy. Also, you dont need to calculate initial velocity. Just use the spring potential energy.

Last edited: Jun 8, 2009
7. Jun 8, 2009

### brunettegurl

so then wld it look like mghinitial+0.5*kx2= mgdsin$$\vartheta$$...where hinitial= 11m ??

8. Jun 8, 2009

correct.

9. Jun 8, 2009

thanks :)