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Work done by several forces

  1. Oct 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Problem # 1
    t72Lb2d.png
    2. Relevant equations
    Work done by several forces

    w=f * d cos theta
    w=mg
    n=w

    3. The attempt at a solution
    I can't figure out how to solve a) and b)
    NOT SURE:
    so my attempt a) f = mg = 30kg*9.8m/s^2 = 294N =magnitute of the force must the worker apply.(NOT SURE)
    b) WofF= f d cos theta = 294N * 4.5m * cos0 = 1323N
    ??????????????????
     
    Last edited: Oct 23, 2016
  2. jcsd
  3. Oct 23, 2016 #2

    kuruman

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    Hi Adrian Pete and welcome to PF.

    Your "relevant equations" show no equations. What do you mean by "work done by several forces"? How many forces do you have acting on the crate? How many of them are doing work?
     
  4. Oct 23, 2016 #3
    5 forces acting on it Sir
     
  5. Oct 23, 2016 #4

    kuruman

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    Can you name these 5 forces and find how much work each one does?
     
  6. Oct 23, 2016 #5
    a) work must the worker apply = :(
    b) work done by the force = :(
    c) work done by friction = Wf = 73.5N*4.5m
    d) work done by gravity = Wg = 294N*4.5m
    e) work done by normal force Wn = 294N*4.5m
    only these Sir :( can't figure out to solve a) and b)
     
  7. Oct 23, 2016 #6

    kuruman

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    First, it seems you have misunderstood the questions (a) and (b). It also seems that question (a) has a typo. I believe the first two questions should read
    (a) What magnitude of force (not work) must the worker apply?
    (b) How much work is done by this force?
    There are only 4 forces exerted by (1) the worker, (2) gravity, (3) normal force and (4) friction.
    It also seems you think that Work = Force × Distance. This is not necessarily true and is not true for some of the forces here. Please look up the definition for work done by a constant force and apply it correctly.
    Finally, you are told that the crate is moving at constant velocity. What must be true for this to be the case?
     
  8. Oct 23, 2016 #7
    force = m * a?
    so a is 0 because it moves at constant velocity?

    so force = 30kg * 0 m/s^2

    a) f = 0N

    so work is w = f d cos theta so,

    w = 0 * 4.5m cos 0

    so

    b) w = 0J = work done by the worker?
     
  9. Oct 23, 2016 #8

    kuruman

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    This is correct. It is not correct that the force exerted by the worker is zero. The worker must exert a force otherwise the crate will not move. Given that the crate is moving and that its acceleration is zero, what must be tru about the forces acting on the crate?
     
  10. Oct 23, 2016 #9
    since there is no acceleration therefore,

    F - fk = ma = 0

    F = fk ?

    so F is basically fk?

    so work done by F is the same as work done by fk?
     
  11. Oct 23, 2016 #10

    kuruman

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    In magnitude. The work done by fk is negative while the work done by F is positive.
     
  12. Oct 23, 2016 #11
    oh okay, because cos 180 is -1. okay Sir. Thank you for the help ^ _ ^
     
  13. Oct 23, 2016 #12

    kuruman

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    That's correct. You are welcome.
     
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