1. Not finding help here? Sign up for a free 30min 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!

Work problem

  1. Nov 18, 2005 #1
    hi,
    Please help me with this problem, thanks.

    Harry pushes a box (that weighs 98 N) with a force of 20 N on ice. After he stops pushing the box is still in motion. The kinetic energy of the box is 20 J and the coefficient of friction is .05. Find
    a) Mass of the box?
    b) Force of friction on the box?
    c) Magnitude of work done by friction to stop the box?
    d) How far the box will slide before stopping?

    I figured out the 1st part.
    F= mg
    98 N = m * 9.8 m/s^2
    m= 10 kg

    The mass of the box is 10 kg.

    I don't know how to proceed further. Please help, thanks.
     
  2. jcsd
  3. Nov 18, 2005 #2
    Part B can be solved using the coefficient of friction and the normal force.
    Part C would be the same as the kinetic energy, because all of the energy must be dissipated.
    Part D would be solved using the Law of Conservation of Energy.
     
  4. Nov 18, 2005 #3
    This is what am comming up for the B part

    Fn - mg -Fp sin(theta) = 0
    Fn = mg + Fp sin(theta)

    mgcos(theta) - Ff = ma

    I dont know the angle theta, what should i do know, thanks.
     
  5. Nov 18, 2005 #4
    I don't see any angle in the original problem you posted. According to the problem, it sounds like it is just being pushed straight, so no angle should be involved. And you don't need the mgcos(theta) - Ff = ma for part B. To solve for friction you only need the normal force and the coefficient of friction.
     
  6. Nov 18, 2005 #5
    I have solved the problem for u, For nr 2 You take Fg*m*0,05 = And you get the Frictionwork. For nr 3 its the same because all energy does not go away it transforms so it must be 20J. And nr 4 it's just to take S = W/F because the defintion for work is W= F*S

    I hope this is right, this is what is what popped up for me anyway.
     
  7. Nov 18, 2005 #6
    Thanks, but wont u think the friction force and displacement are in the opposite direction so the angle between them will be 180 degrees.
    W= F*d cos 180. What do u think, thanks for looking at my problem.
     
  8. Nov 18, 2005 #7
    Thanks for your help.

    Ff = coeff of friction * Fn
    = .05 * 98 N
    = 4.9 N

    For the work done by friction force
    W(f) = Ff * d cos 180
    since the friction and displacement are in the opposite direction,so the angle between them is 180. Now what should i do for W(f). I have two unknowns W(f) and d, thanks.
     
  9. Nov 18, 2005 #8

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    Actually, you can figure out what W(f) is from the problem.
     
  10. Nov 18, 2005 #9
    Will W(f) will be -Ff, thanks
     
  11. Nov 18, 2005 #10

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    The box starts out in motion (with 20 Joules of Kinetic Energy) and then comes to a stop. What forces are doing work on the box?
     
  12. Nov 18, 2005 #11
    drawing a free body diagram might help
     
  13. Nov 20, 2005 #12

    The forces doing work on the box are Ff (frictional force), Fn (normal force), gravity (mg), and Fp (force of pus). When the box is in motion it has kinetic energy. Am not getting this one sorry.
     
  14. Nov 20, 2005 #13
    is there really the force of the push on the box? check the question again.
     
  15. Nov 20, 2005 #14

    lightgrav

    User Avatar
    Homework Helper

    Force describes a *condition*, visible in a snapshot diagram (FBD).
    Work is something that occurs (is DONE) during a process -
    the process changes the starting condition into the ending condition.

    The Force component *parallel to* the displacement does Work
    as the displacement is accomplished (W = F.dx)
    This means that Energy is transferred from one object to another
    (thru the contact area) as that contact point moves.

    Where does the block's KE end up?
     
  16. Nov 21, 2005 #15
    The force of push is there on the box intially. After the box begins to move there is no push.
     
  17. Nov 21, 2005 #16
    The box's KE ends up when it comes to stop.
     
  18. Nov 21, 2005 #17
    yeah, so that means that there's 20 Joules of kinetic energy initially, and the work done by friction decreases it all to zero. So, the [itex] -W_f = KE [/itex]. You can solve for the displacement of the box by using [itex] W_f = F d cos(Theta) [/itex]

    the direction doesnt matter, since the question asks for a magnitude
     
    Last edited: Nov 21, 2005
  19. Nov 27, 2005 #18

    Thanks for ur help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work problem
  1. Work problem (Replies: 2)

  2. Problem on Work (Replies: 3)

Loading...