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Work with velocity and force

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A box of mass 49 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0.60. A woman pushes horizontally against the box with a force of 628 N until the box attains a speed of 3 m/s.

    What is the work done by the woman on the box?


    What is the CM-work done by the friction force on the box?

    2. Relevant equations

    W=F (delata x)
    W(f)= uN (delta x)

    3. The attempt at a solution

    Since work is in joules, I tried calculating it by multiplying force by velocity squared but that gave me a wrong answer. If I had the right answer for work I'd solve for delta x and use that to solve for work of friction
  2. jcsd
  3. Oct 10, 2013 #2
    Try to think Kinematics: You know the mass, you know that it starts from rest, and you know the box's final speed.
  4. Oct 10, 2013 #3
    Also CM-work is center of mass right? not centimeter?
  5. Oct 10, 2013 #4
    I believe so, yes.
  6. Oct 10, 2013 #5
    vf^2 = vo^2 + 2(a) d? seems to be what I'm looking for maybe. Do I use F=ma to splve for a? so 3^2 = 2 (d) 628/49. making d = 441/1256? so W= 628 (441/1256) = 220.5
  7. Oct 10, 2013 #6
    I believe you can use f=ma to solve for acceleration. From there, use a kinematics equation to solve for distance in the x direction. Then you can multiply that by the force to solve for work.
  8. Oct 10, 2013 #7
    But that's the KE, they're not the same are they?
  9. Oct 10, 2013 #8
    So delta KE is 220.5, work by woman 220.5, and work by the force of friction = (u)mg (d) = .6(9.8)49( 441/1256) = 101.1631... ? Is that right?
  10. Oct 10, 2013 #9


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    That's why dimensional analysis is so handy.

    Energy and work have the same derived units: W = (M*L^2/T^2)

    If you apply dimensional analysis to your original supposition that W = F*V^2,
    you would see that F = M*L/T^2 and V^2 = L^2/T^2, so W = M*L^3/T^4,
    which isn't even close to the correct W = M*L^2/T^2
  11. Oct 10, 2013 #10
    Yeah, I didn't use that supposition. I used vf^2 = vo^2 + 2(a) d instead to find distance... I just want to know if my answers for work of the woman and of the friction force are correct?
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