# Homework Help: Spring pushing block down and up hill problem

1. Mar 13, 2010

### bc2man

1. The problem statement, all variables and given/known data
3.5 kg box is launched from a spring that has spring constant 1200 N/m. The spring is initially compressed by an amount 1.7 m. The box slides along the frictionless track shown below and into a second spring that has a spring constant 900 N/m. It compresses the spring by an amount 2.1 m before turning around. What is the height of the second spring?

2. Relevant equations
F=ma (for 3 different phases)
F=-kx (hookes equations)
V_f^2 = V_0^2 + 2a(delta_y)
mgh=delta_KE

3. The attempt at a solution

I feel like my attempt at a solution went too smoothly, so i'm suspicious I did something terribly wrong somewhere, or forgot to account for something.

Basically I started at the blocks starting position. Using Hooke's law, I got F=-kx, and F turns out to be -2040 N... (-1200N/m / 1.7m). I then do the same thing for the other side where the spring compresses at the end of the blocks path. I get F= -1890N ... (-900N/M)(2.1m).

So there is 150 N lost in the system... this must mean 150N is lost when its going up along the higher slope on the other side?

I then attempt to find V_final using mgh = (1/2)mv_f^2 - (1/2)mv_i^2. But v_i^2 is just 0 so...

mgh= (1/2)mf_f^2.

I plug in numbers, do the algebra and v_f comes out to be 11.46 m/s ...

I THEN use V_f^2 = V_0^2 + 2a(delta_y)

Same story... plug in numbers and do algebra and I get delta_y to be 1.53m.

Now I tack on the starting height plus this found height. So I get 6.7m + 1.53m and I end up with the final answer of 8.23m

Does this all sound right?
I dont have a solution to the problem so I wanted to check with the pro's on this forum.
Thanks for any help!

2. Mar 13, 2010

### Staff: Mentor

Force is not something that is 'lost' or conserved. Hint: What is conserved as the block slides from one hill to another?

3. Mar 13, 2010

### bc2man

I'm not sure which equation to use then?
I have something written in my notes to the effect of:
mgy_i + (1/2)mv_i^2 = mgy_f + fk(s)

I'm not sure what s stands for but it cancels anyway (fk is 0...system is frictionless).
the (1/2)mv_i^2 term also cancels as v_i = 0
But then I'm just left with my starting height of 6.7 m on the other side?

I guess this is wrong too?

4. Mar 13, 2010

### Staff: Mentor

That equation is incomplete. Add spring potential energy to the mix.

5. Mar 13, 2010

### bc2man

Ok... so I KNOW spring potential is (1/2)kx^2 but I'm confused on how to incorporate it into the above equation...
would it be something like mgy_i + (1/2)kx^2 = mgy_f ?
Or is it two (1/2)kx^2 terms because of the 2nd spring?
I tried this out and I got a rediculously high final delta_y value...

6. Mar 13, 2010

### godtripp

Well you should know that
$$KE_{i} + U_{i} = KE_{f} + U_{f}$$

This is equivalent to saying that energy is conserved. So Kinetic Energy initial + potential energy initial is equal to Kinetic Energy Final + Potential Final.

A spring is just potential energy. Energy is stored in the spring, when you let the spring go that energy turns into kinetic energy. So if you start on a spring, you just have an initial potential from the spring.

If you are in motion and run into a spring, your kinetic energy transforms into potential energy. So you have a final potential energy from the spring.

Hope this helps, and let me know if you need anything else clarified.

7. Mar 13, 2010

### bc2man

so we're starting with just spring potential??

I take it the equation might look something along the lines of

(1/2)kx^1 (of spring 1) = (1/2)kx^2 (of spring 2) +mg(h-6.7m) ??

8. Mar 14, 2010

### godtripp

No, not JUST spring potential. But you must include spring potential.

You start out not moving, you don't have any kinetic energy initially.
The question says after the block runs into the second spring it compresses 2.1 m before being shot out the other way. So your final kinetic energy is also 0 for that split second.

Understand why?

So since kinetic energy initial and final is equal to 0 you're left with.

$$U_{i} = U_{f}$$

Your potential initial and final should be your TOTAL potential at those points so remember you have potential from gravity at each of those points as well as potential from your spring.

so let's build equations for potential initial.

$$U_{i} = mgh_{i} + \frac{k_{i}x_{i}^{2}}{2}$$

where $$h_{i} = 6.7$$ m and $$x_{i} = 1.7$$ m and $$k_{i} = 1200$$ N/m

So, you can easily build an equation for your final potential

$$U_{f} = mgh_{f} + \frac{k_{f}x_{f}^{2}}{2}$$
remember, at the second position you have a different spring constant (k) and the spring is compressed a different distance (x). you don't know the final height so leave that as your variable h and solve for it. There's your answer.

I assume this is one of your first times attempting energy problems, so if you don't understand why I chose to do anything that I did in these steps let me know so I can explain them, that way you can approach your other problems with more ease.

9. Mar 14, 2010

### bc2man

Thank you soo much god tripp and Doc... I think I'm on the right path to understanding these questions now.

In terms of the answer, I got 3.04m for the h value. This has to be right, although the picture doesn't represent it too well...I guess my professor is trying to throw us off with radical not to scale type diagrams.
Thanks again!

10. Mar 14, 2010

### godtripp

I haven't checked for myself if your numerical answer is correct or not.

Don't think of any picture in energy problems to be necessarily to scale, think of h as a changing variable, so it could easily stretch way up or even stretch below the point you call h=0.

It might help you in other problems to know that you can call any point h=0. Let's assume the question asked for how much above or below the initial position is the final block. Then it would be easier to say that h=0 is at the initial point, so your initial potential from gravity is = 0. and the variable h in your final potential from gravity would be the distance above the initial if h turns out to be positive, and below the initial if h is negative