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Homework Help: Kinetic Energy Question

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A crate of mass 10.00kg is pulled up a rough incline with a n initial speed of 1.50m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20.0 degrees with the horizontal. The coefficient kinetic friction is 0.400, and the crate is pulled 5.00m.
    What is the change in kinetic energy if the crate?

    2. Relevant equations

    3. The attempt at a solution
    This what i have done but i cannot seem to get the right answer:

    To find velocity:
    W = 0.5mv^2 + Frictional force * d
    500 = 5v^2 + 184
    v= 7.98m/s

    To Find change in kinetic energy:
    K = 0.5mv^2
    K = 5 * 7.95^2
    K = 316

  2. jcsd
  3. Mar 21, 2010 #2

    Filip Larsen

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    It may help to write up a diagram showing all the forces that act on the mass during its slide up the incline. Since you are interested in distances and forces parallel to the incline you should be able to project all the forces unto this direction, think about if they are constant or not, and then apply your energy equations (hint: increase in kinetic energy of the crate is the sum of the work (with sign) done by the pulling force, friction and gravity; kinetic energy then determines speed).
  4. Mar 21, 2010 #3
    Hey Pay, I'm a little confused. What is the answer you are looking to get?

    First, your calculation of the final velocity doesn't depend on the initial velocity. If you needed the velocity it would come from this,

    [tex]K_{f}[/tex] = [tex]K_{i}[/tex] - [tex]f_{k}[/tex] d + W

    Secondly, for your KE calculation you are not computing a CHANGE in energy. Your calculation is the energy of the block at the instant it is moving 7.95 m/s. (I don't know where you got that velocity?)

    But, I think you get the answer from the less explicit form of the work-kinetic energy thm:

    [tex]\Delta[/tex]K = - [tex]f_{k}[/tex] d + W

    [tex]\Delta[/tex]K = -[tex]\mu_{k}[/tex]mgCos([tex]\theta[/tex]) d + F d

    Here's the confusing part, when I did that calculation quick it came out to,

    [tex]\Delta[/tex]K = 315.82 J.

    The question only asks for the change in kinetic energy and you seemed to have got that somehow ; )
  5. Mar 21, 2010 #4
    no, 315.82 is not the correct answer according my book's answers.

    I worked out that to get the change in kinetic energy you need to add up all the forces exerted by the crate to get the correct answer (according to what filiplarsen said).

    so my final answer is 148 Joules.
  6. Mar 21, 2010 #5
    You can't add up forces to get a change in kinetic energy. Filplarsen correctly said you are dealing with the sum of forces applied over distances. Or work. A change in kinetic energy is work and that's what the problem, as written, is asking for.

    Forces on the box that do work over the 5.00 meters:

    Pulling force = 100 N. Friction force = u_k*N = -36.8 N.

    Net force = 100 +(-36.8) = 63.164 N.

    Again, if I multiply that by 5.00 m. I get 315.82 Joules. Is 148 J. the answer in the book? Please show quickly how you got an answer and what you are aiming for. Otherwise, I can't figure out where you got 148 J.
  7. Mar 21, 2010 #6

    Filip Larsen

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    I also get 148J. Remember there is a third force at play here.
  8. Mar 21, 2010 #7
    Yep, I left out gravity, thanks guys. 148 J. is definitely right.
  9. Mar 22, 2010 #8
    ok i got another question related to this problem.
    What is the speed of the crate after being pulled 5.00m?

    This is what i have done, but unsure if it is correct. Answers says it 5.65m/s, however i got 5.44m/s




  10. Mar 22, 2010 #9

    Filip Larsen

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    You are on the right track but have missed something. Try read the problem text carefully.
  11. Mar 22, 2010 #10
    hmm.. is something to with the change in Kinetic energy?
  12. Mar 22, 2010 #11


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    148 = Kf - Ki= 5(vf^2 - vi^2)
    Now find vf.
  13. Mar 22, 2010 #12

    Filip Larsen

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    The 148 J is the change in kinetic energy, but that is not necessarily the final value of the kinetic energy (hint: did the crate had any kinetic energy to start with?)
  14. Mar 22, 2010 #13
    Did you get 5(vf^2 - vi^2) equation from v^2= u^2 +2as????
  15. Mar 22, 2010 #14


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    Work done = Change in the kinetic energy
    148 = 1/2*m*(vf^2 - vi^2)
  16. Mar 22, 2010 #15
    ok, i get it now thanks
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