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Questions re. KE/Work

  1. Oct 17, 2011 #1
    Hey everyone, I'm really lost with my physics homework!



    A ball of mass 250 g is dropped 5 m onto a spring
    -how much KE wil the ball have when it hits the spring?
    -how much work will be done as the spring is compressed?
    -If the spring is 250 kN/m, calculate how far the spring will be compressed?




    3. A ball of mass 100g is hit vertically upwards with a bat. The bat exerts a constant force of 15 N on the ball and is contact with it for a distance of 5cm.
    -how much work does the bat do on the ball?
    -how high will the ball go?


    I've tried setting up equations, but I can't seem to get answers that are realistic. It's very possible that I'm not using the right formulae, so maybe I should start there?

    (Here's my working thus far: I'm completely stuck on 1a, but for 1b I converted the grams to newtons so..
    w=f*d
    w=2.5*5
    w=12.5 J (but i'm not sure)
    again, c confuses me a lot

    for 2a, I did
    w=15*.05=.75 J, but I'm not sure where to go from there!)
     
    Last edited: Oct 17, 2011
  2. jcsd
  3. Oct 17, 2011 #2
    You should show your steps so we can help you more appropriately.:)
     
  4. Oct 17, 2011 #3
    Sorry, thanks for your response!!

    I'm completely stuck on 1a, but for 1b I converted the grams to newtons so..
    w=f*d
    w=2.5*5
    w=12.5 J (but i'm not sure)
    again, c confuses me a lot

    for 2a, I did
    w=15*.05=.75 J, but I'm not sure where to go from there!
     
  5. Oct 17, 2011 #4
    hmm okay I see where you are having some trouble. The ball being 5 metres off of the ground represents the fact that it has a certain amount of potential energy. At the instant before the ball hits the spring, all of that potential energy has been converted into kinetic energy. So all that question is asking you is how much potential energy did the ball start off with? The work done is basically the energy put into the spring. And you know this value already. And part c isn't too hard. There is an equation for energy inside a spring. This equation is Energy = 1/2 k * x^2. So basically this is giving you a relationship between energy and displacement, using the spring constant value.
     
  6. Oct 17, 2011 #5
    Sorry, I'm confused. How do I know how much energy the ball started off with? And does that make my answer for B incorrect as well?
     
  7. Oct 17, 2011 #6
    I haven't done any of the calculations yet. Do you know the equation to find the potential energy of a mass at a certain height above the ground?
     
  8. Oct 17, 2011 #7
    Not sure...I did: PE=m*g*h
    --> .25*9.8*5 = ~12.5, but that's the same as what I got for B!?
    Also, could you possibly help clarify the "how high does the ball go?" aspect of the 2nd question?
     
  9. Oct 17, 2011 #8
    sounds like you got B right. And that is the formula, so you have A right as well!

    For the second question you can use the same energy related ideas that you used for the first question. From your first post you use the equation work = force * distance, you can use this equation again.

    Your work will equal the initial kinetic energy of the ball
    the initial kinetic energy of the ball will be equal to the final potential energy of the ball(the same as in question 1).
    Once you have the final potential energy, you know how to relate potential energy to height.
     
  10. Oct 17, 2011 #9
    Thank you so much!! I think I (somewhat) understand what to do!
     
  11. Oct 17, 2011 #10
    Haha awesome! Good luck to you. If you have any other questions you can Personal Message me if you'd like, or posting here might flag me to come back.
     
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