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Can SomeOne Check My Work?

  1. Nov 7, 2005 #1
    A 45Kg block slides down a frictionless incline 1.5m long and 0.91m high. A worker pushes up against the block, parallel to the incline, so that the block slides down the incline at a constant speed.

    (a)what is the magnitude of the workers force?

    (b)How much work is done on the block by the workers force?

    (c)The Gravitational force?

    (d)The normal force on the block from the surface of the incline?

    (e)The Net Force on the block.

    m = 45Kg
    d = 1.5m
    Theta = 37.35
    a = 0m/s^2

    I drew a F.B.D and came up with the following:

    (-F) - mgsin(theta) = ma = 0
    -mgsin(theta) = F
    F = -267.5N

    Wa = Fd
    Wa = -267.5N(1.5m)
    Wa = -401 J

    Wg = -Wa
    Wg = 401 J

    N - mgcos(theta) = 0
    N = mgcos(theta)
    N = 350.6 N

    F(total) = ma
    F(total) = 0
  2. jcsd
  3. Nov 7, 2005 #2
    (a) Correct approach but what is the positive direction?
    (d) Careful. Which direction does the normal force act. Does it have any component parallel to the displacement of the box?
  4. Nov 7, 2005 #3
    Positive Direction = up the ramp

    The normal force is perpendicular to the ramp isnt it?..it doesnt have any component parallel to the displacement i dont think.

    So i'm guessing that my answers are wrong?
  5. Nov 7, 2005 #4
    If positive is up the ramp, then your answer means the guy is pushing down the ramp (you gave a negative answer).

    So if it has no parallel component, did it displace the box? Waht does that mean about the work done?
  6. Nov 7, 2005 #5
    but he must be pushing up the ramp to stop the box from accelerating right?

    the normal force wouldnt displace the box i dont think....

    Not too sure though..I dont know what to do
  7. Nov 7, 2005 #6
    Yeah, but if he's pushing -267 ni the UP direction it means he's pushing 267 in the down direction. The mistake you made is in the first line of your work. Redraw your FBD.

    Right it doesn't, if the Force doesnt cause any displacement, it does no work.
  8. Nov 7, 2005 #7
    i think i see it...maybe..

    I drew the force pushing downwards..when it should have been upwards..soo..

    F - mgsin(theta) = ma = 0
    F = mgsin(theta)
    F = 267.5N

  9. Nov 7, 2005 #8
    You goti t.
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