I hope that helps. How far will Lisa compress the spring?

[octahedron]

Homework Statement

A new event has been proposed for the Winter Olympics. An athlete will sprint 100 m, starting from rest, then leap onto a 20 kg bobsled. The person and bobsled will then slide down a 50 m long ice covered ramp, sloped at 20°, and into a spring with a carefully calibrated spring constant of 2300 N/m. The athlete who compresses the spring the farthest wins the gold medal. Lisa, whose mass is 40 kg, has been training for this event. She can reach a maximum speed of 12 m/s in the 100 m dash.

How far will Lisa compress the string?

http://img258.imageshack.us/img258/5415/p1056ra6.gif

Homework Equations

(0.5)*k*x_i^2 + (0.5)*m*v_i^2 + mgh = (0.5)*k*x_f^2 + (0.5)*m*v_f^2 + mgh

The Attempt at a Solution

0 + 0 + (40+20)(9.81)(50sin(20)) = 0.5*(2300)*(x_f)^2 + 0.5(60)(12)^2 + 0

after calculating everything, I'm getting

x_f = 2.23 meters

while the answer in the back is 3.2 meters. There's likely something in my solution that I missed. Could anyone point it out for me?

 

[Supernats]

Your equation in (2) looks to be saying x_i = x_f. That doesn't seem right.

Edit: Also, you didn't square x_i in your solution.

 

[octahedron]

Corrected! Apologies for doing this again.

Edit: Also, you didn't square x_i in your solution.

I assumed that the spring is relaxed, thus x_i = 0. Is this flawed?

 

[Supernats]

Corrected! Apologies for doing this again.



I assumed that the spring is relaxed, thus x_i = 0. Is this flawed?

No, it isn't. I was looking at the wrong term.

 

[Supernats]

I'm having problems seeing how you can say initial kinetic energy is zero here. If you're using the combined mass for mgh, I think you have to use 12 m/s for v_i.

 

[octahedron]

Don't I still need to use that same velocity to find the compression distance in the RHS? If I use it initially, both KE's will cancel.

Edit: I do see your point that there does exist a v_i in the problem, which makes KE_i exist. Escapes me how to find it though.

 

[Supernats]

When the spring is at max compression, velocity is zero. If it weren't, you would still have energy to be putting into the spring. Basically, everything goes into the spring.

 

[octahedron]

Gotcha. So KE_i + PE_i = PE_spring_f ?

(.5)(20+40)(12)^2 + (20+40)(9.81)(50sin20) = .5(2300)(x_f)^2x_f = 3.54 meters. Does this sound about right? A little higher than the projected book value.

 

[Supernats]

Yeah, I get the same value. I don't know, everything looks right.

 

[eorstad]

I had the biggest issue with this problem until I realized that I was forgetting about momentum at the beginning of the problem!

m x v_i = (m + M) v_f

40kg x 12m/s = 60kg x v_f

v_f = 8 m/s

now you should get the correct answer!

 

[Cliffdizzl]

Basically, you want to find the energy right before she gets to the spring. This is a two part problem:

1) find vf in the equation .5m(vi)^2 + mgh = .5m(vf)^2 + mgh
("h" is zero on the right side at the bottom of the hill, so actually solve this equation for vf:)
.5m(vi)^2 + mgh = .5m(vf)^2

Remember, "vi", although it appears to be 12 m/s, is actually 8 m/s in this problem because you need to take momentum into consideration (as discussed above). if lisa (40kg) runs at 12 m/s and jumps onto the sled of 20 kg, then take the momentum (40kg*12 m/s) and divide by the new total mass (60 kg) to get 8 m/s.

Now, you should solve for "vf" in the previous equation and get 19.97 (or just 20) m/s


2) Now we can solve the next equation: the kinetic energy (.5m(vf)^2) right before Lisa hits the spring is equal to the spring's elastic equation .5K(xf)^2

Solve for "xf": .5m(vf)^2 = .5K(xf)^2 (and remember, "vf" is 20 m/s)

after all this, you should get about 3.23 m, as the answer in the first post states