1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Springs and potential energy

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data
    A 2kg block is pushed against a spring of force constant k=400 N/m, compressing it .22m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with a slope of 37 degrees. a.) what is the speed of the block as it slides along the horizontal surface after having left the spring? b.) How far does the block travel up the incline?

    2. Relevant equations
    x=x0 + V0t+ 1/2 at^2
    3. The attempt at a solution
  2. jcsd
  3. Mar 31, 2015 #2
    Sorry I can't give you the answer straight out.

    Think about conservation of energy. Starting with part a, where would the speed come from?

    If this is indirect it's because that's how it has to be.
  4. Mar 31, 2015 #3
    i think i figured out the velocity in part a, but i am having a lot of trouble finding the acceleration to find the distance.
  5. Mar 31, 2015 #4
    Ok. So it has speed and now it is going up an incline.

    Let me see how I should talk about this:

    It has left the spring so there is no longer a force applied by the spring on the block. You might want to think about what forces are left to act on the block; the fact that the block is moving up an incline is a big clue (what force fights gaining height?).

    Please use energy for this. You could use the x = x0 + vot + 1/2 a t^2 in theory but I would highly advise against it. This question was meant to show the simplicity of conservation of energy (probably in the energy chapter in your book). Think about how energy is transferred as the block goes up the incline.

    Sorry this is sort of roundabout.
  6. Mar 31, 2015 #5
    that's okay. See i was thinking i could use V^2=V0^2+2ad and plug in 0 for the velocity, plug in the initial velocity that i got in part a, and plug in g for acceleration and solve for d. But since at this point it would not be moving, then the acceleration would also be 0 too. So, I am not sure
  7. Mar 31, 2015 #6
    Below is a justification for why the block in this problem is always accelerating even once it has stopped due to gravity. You may want to talk to your teacher for a more personal explanation since that might be more intelligible:
    Ok when it stops the acceleration due to gravity will still act on the block just so you know. Acceleration is change in velocity; even if something has stopped it still could be gaining or losing velocity. You can think about the fact that the block will start sliding back down the incline (this is not part of the question though). The velocity goes from zero to something negative and thus there is a change in velocity and thus there is an acceleration.
    I don't want you to generalize this too much though because if the block was on a flat ground it would not be accelerating due to gravity; this is because the ground is very effectively pushing back up on the block while gravity is pushing down so there is no net (total) force and therefore there is no net acceleration.

    Back to the energy part:

    That equation looks good. Just remember that d is the distance in the same direction of the acceleration that the object has moved (ask me to elaborate if you need me to).

    This origin of that equation comes straight from energy conservation. F * (d) = change in kinetic energy = 1/2 m vf^2 - 1/2mv0^2. This is where vf is final velocity and vo is initial velocity. This is how you arrive at the equation you just proposed (if you're interested) (note that for your equation you just multiply by two on both sides, divide by m on both, and add vo^2 to both sides (sorry for edit: I didn't look at your version very carefully)).
    Last edited: Mar 31, 2015
  8. Mar 31, 2015 #7
    Just a note that I've been worried about:

    I don't want to confuse you; nonetheless I should clarify this because I don't want you to be confused when it comes time to take a test.

    It happens that in the problem you were dealing with the block was accelerating at the top of the incline.
    But, even if the block was not accelerating at the top of the ramp (let's pretend someone held it in place or whatever you want) the energy lost for the block to get to the top of the ramp is still the same. This is because it was still "fighting" gravity all the way up. If this is confusing feel free to ask for clarification. But mathematically the way to look at it is: change in energy (gain or lost) = F * distance (the distance has to be on the same axis as the force). For example energy change due to gravity occurs when you go up or down, not side to side.

    You can also use your equation. Distance operates in the same way. That case is specific to potential energy changing into kinetic or kinetic into potential (hope you know what those words mean).

    I hope this was not confusing.
  9. Mar 31, 2015 #8
    no, for the most part it does make sense.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted