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Help with a few problems.

  1. Apr 13, 2004 #1
    Sorry...I hope I'm not annoying anyone. I really have tried to figure out how to do these, but I jsut can't get it. I hate jsut throwing problems out there, but I dont really have a better way of doing it.

    "Karmen is playing pool, and strikes a 0.25 kg ball, making it go 1.5 m/s. If Karmen used the force of the cue on the ball with 20 N, over what distance did it act?"

    I think I should use W=Fd at some point, but still, that gives W=20d which still doesnt get anywehere.

    "A 30 kg shopping cart is on a 2m hill and rolls down to hit a stump at the very bottom. At impact, a .25 kg can flies and hits a car with an average force of 490N. How deep is the dent?"

    I think we are supposed to assume the metal did work in stopping it. I really don't know!
     
  2. jcsd
  3. Apr 13, 2004 #2
    Use Work Energy Theorem
    [tex] W = \Delta K [/tex]
     
  4. Apr 13, 2004 #3
    What is K in that equation?
     
  5. Apr 13, 2004 #4
    Kinetic energy. More commonly denoted T or KE.

    cookiemonster
     
  6. Apr 14, 2004 #5
    Okay...so if KE=.5mv^2, then...
    KE=(.5)(.25)(1.5)^2
    KE=0.28125 J, right?
    Or is it even in Joules? I'm really confused about the units of measurement.

    Anyhoo...
    Work=.28125
    Work=Fd
    20d=.28125
    d=0.0140625 m, right?

    How in the world do we figure the can's velocity? Really. The fact that the cart rolled down the hill would not mean anything about the cans velocity unless given the hill's angle, right?

    If not, how do I find the velocity of something at the end of a hill of x height?
     
  7. Apr 14, 2004 #6

    enigma

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    Staff Emeritus
    Science Advisor
    Gold Member

    You use the same basic idea:

    Conservation of energy.

    Assuming no friction (which I'm guessing is the case for this problem even if it didn't state it specifically), It doesn't matter what angle the cart rolls down the hill. As long as the hill is the same height, you'll end up with the same velocity at the bottom regardless of angle (try it!)

    mgh=1/2 mV2
     
  8. Apr 14, 2004 #7
    mgh=(30)(9.8)(2)
    SO the cart has potential energy of 588 kilogram-meters per second squared-meters.
    And we are saying that the can receives the full amount, in kinetic energy?

    588=.5(.25)v^2
    v=68.59 m/s!

    Wowzas. Anyhoo. Delta-KE=Work
    588=490d
    d=1.2 meters!

    I really doubt that. Where'd I mess up?
     
  9. Apr 14, 2004 #8

    Doc Al

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    Staff: Mentor

    No, but both the cart and the can fall the same distance. Figure out the ΔPE of the can--that will be the KE of the can at the bottom of the hill, when it flies out of the cart.
     
  10. Apr 14, 2004 #9
    Alright, I see...so it's potential energy will simply be .25/30 of the whole cart's, right? That'd be 4.9 units of energy.

    So...4.9=490d
    d=0.01 m or 1 cm.

    Sounds great! Thank you all!
     
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