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Work energy problem

  1. Jan 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A 30.0 kg box sliding at 5.00 m/s on a rough surface is brought to rest by 20.0 N of friction. What distance does the box slide?

    2. Relevant equations
    KE = 1/2*m*(v^2-v^2)

    3. The attempt at a solution
    I know that the change in velocity is 5.00 m/s since it is going to rest, I understand the weight and how it works into the equation... but where the Newtons come into play and how I use this to find the change in distance is where I am lost.

    I know that Joules are Newton-meters (N*m), but I'm not too sure on converting.

    If I put 30.0 kg and 5.00 m/s into the above equation I get KE = 0.5*30.0 kg*5.00 m/s = 375 J = 375 N*m
  2. jcsd
  3. Jan 3, 2008 #2

    Shooting Star

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    So, where's this energy gone?
  4. Jan 3, 2008 #3
    I would use Vf^2 = Vi^2 -2AD and keep in mind that A = F/M
  5. Jan 3, 2008 #4
    Yah, to ring the same bell Shooting Star did, when the box starts it has 375 joules of energy and when it stops it has 0.

    I'm sure you realize the friction took the energy away. Conveniently you're given the force of friction. Mayhap you know a way to relate the work done by friction with the force and distance travelled?
  6. Jan 3, 2008 #5

    Shooting Star

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    Not necessary here.
  7. Jan 3, 2008 #6
    Yea I just looked at the problem again, solvable with work-energy
  8. Jan 4, 2008 #7
    Since I'm trying to figure this out, allow me to type what I'm thinking here... hopefully you guys could point out where I'm going wrong:

    So since friction has absorbed the energy that was being used to move the block, my goal is to figure out the distance covered by whatever work the friction did to bring the object from whatever velocity it was going to rest. Does this mean that I can use the W=Fs formula and solve for s to find the distance?

    since F=ma, a=F/m as cryptoguy said... or what about s = Wma...

    but how do I use acceleration? am I supposed to take the derivative of velocity somewhere? I guess if I could make a formula for its change in acceleration I could integrate it and get a formula for its position, then find the formula's x-intercept... but I don't have one of those.

    But it can't be as simple as W J = 20 N * s

    Well I have a mass and a velocity... I can find momentum then:
    p = 30000 g * 5.00 m/sec
    dp / dt = 0 + 30.0 kg * [ d(5.00 m/sec) / dt ] = F

    so the derivative of 5.00 m/sec is my acceleration. Does that mean that a = 5.00 m/sec/sec ? That doesn't sound right to me.

    But since I have my acceleration, I can now say that W = ma * s => s = W / ma

    but I don't know what the work is...

    if I say that a = F/m like cryptoguy said and replace F with 20 N and m with 30.0 kg, then I have my acceleration! So 20 N / 30.0 kg = 2/3 N/kg = a = 2/3 m/sec/sec.
    Does this mean that N/kg is the same as meters per second squared? That doesn't sound right either.

    Wait, 1 N = 1 kg * m / s^2... so yeah kg*m/sec^2*kg = m/sec^2 so yeah it works out!

    a = 2/3 m/s^2

    Now I need to find out the W, so that I can solve for the displacement.

    W = ( 30.0 kg * 2/3 m/s^2 ) s

    s = ( 30.0 kg * 2/3 m/s^2 ) / W = 20.0 N / W kg*m^2/s^2

    So how do I find the work W?

    Perhaps I'm thinking too hard.
    Last edited: Jan 4, 2008
  9. Jan 4, 2008 #8

    Shooting Star

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    You are. Work done against friction = 20*distance.
  10. Jan 4, 2008 #9
    Yah it's a two line problem of which you successfully did one line

    Boxes initial kinetic energy = 375J

    Work done by friction = 375=F*d, you're given the force

    If you wanna be technical, I guess the work done by friction is -375J, and the force is also negative(making distance still positive)

    But shooting star summed it up, I'm primarily posting to let you know your explanation made me laugh and my head explode at the same time
  11. Jan 4, 2008 #10

    Shooting Star

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    That's not a very nice thing to say. He's a newbie who evidently is doing this sort of problems for the first time.
  12. Jan 5, 2008 #11
    Well I didn't mean it to be mean, I'd think it was funny if I did it too
  13. Jan 5, 2008 #12
    I'm not insulted, it's alright.

    I took physics in high school, and now I'm trying to learn it with calculus, now that I've completed the class. Somehow I got the impression that plug-and-chug is not how you're supposed to do this stuff when calculus is brought into the equation. Maybe I'm mistaken.

    But just so I'm clear on how this problem was done so easily:

    When I found the 375 J for the Kinetic Energy, the

    KE = 0.5*30.0 kg*5.00 m/s = 375 J = 375 N*m

    part, I thought that that was the only equation I needed to work the problem, which lead me to think about velocity and position and all that crazy math crap, which really was the long way to do it I guess.

    When you guys mentioned F=ma and W=Fd, I was still thinking along the lines of acceleration and antiderivatives because I had the impression that the answer HAD to have some kind of calculus thing in it, so I went off trying to integrate a bunch of stuff... when really this was just a basic plug and chug thing.

    Clearly I was thinking too hard. I appreciate the help, and don't hesitate to call me retarded or whatever if I make any mistakes.

    A couple more things:

    When blochwave says
    and I solve for d, does this mean that since 375 J is the work done by the friction, that the F is the Force done by the friction, and the d is the distance over which the friction is applied?

    If that's true, then how could I figure out this problem if, instead of coming to rest, its speed decreases from 5.00 m/s like in the question to 2.00 m/s, and then asks me what distance it took for the friction of 20.0 N to slow it down to that speed?

    It would be the same kind of problem, but with a different KE, right?

    KE J = 0.5 * 30.0 kg * [ (5.00 m/s) ^2 - (2.00 m/s) ^2 ]

    KE J = 20.0 N * d

    KE J / 20.0 N = d

    So is the Work done by the friction to slow the 30.0 kg object the Kinetic Energy of the friction... or is the Work equal to the Kinetic Energy of the object because it is the one slowing down and not the friction? Is Work done by something = Kinetic Energy of that something in all cases or just this one because it meets some crazy, convenient condition?
    Last edited: Jan 5, 2008
  14. Jan 5, 2008 #13
    Everything you calculated is correct.

    Friction is not an object so it doesn't have any kinetic energy, but it is a force so it does work. It does work on the object and causes it to lose kinetic energy. Where does it go? Friction is a non-conservative force so the kinetic energy the object loses goes into useless heat.
  15. Jan 6, 2008 #14
    So anything at all can do work? And in this case, the friction does the work because it forces an object to slow over a certain distance, meanwhile the object just relaxes while the friction does all of the work for it?
  16. Jan 6, 2008 #15
    Well not anything can do work, only forces. In this case, friction is the force acting on the object.
  17. Jan 6, 2008 #16

    Shooting Star

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    Work is said to be done when there is a force acting on an object and the object is being displaced.
  18. Jan 6, 2008 #17

    Shooting Star

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    Hi micaele, I am sure that you have got the value of the distance reqd in the original problem. For further conceptual clarification about work done etc, I believe it's best to start a new thread, preferably in the General Physics forum. That'll benefit you.
    Last edited: Jan 7, 2008
  19. Jan 6, 2008 #18
    I understand — thanks. If I get stuck anywhere else I'll be sure to make a new thread.

    Thanks for your help everybody.
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