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Homework Help: HELP: Work Done

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A box of weight 200N is pushed so that it moves at a steady speed alog a ramp, through a height of 1.5m. The ramp makes an angle of 30° with the ground. The frictional force on the box is 150N while the box is moving. What is the work doe by the person?

    2. Relevant equations

    3. The attempt at a solution
    please explain this one.. thanks...
  2. jcsd
  3. Sep 9, 2012 #2


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    Gold Member

    What have you tried? The box moves at steady speed so its change in kinetic energy is zero. There are two forces acting on the box which tend to transfer energy away from the box. THe person must put this energy in to allow the box to continue to move with steady speed.
  4. Sep 9, 2012 #3
    Lets solve this one

    now there is no acceleration in the block right??

    So as velocity is constant, there is no force

    So the forces must cancel out...

    hope now u get it
  5. Sep 9, 2012 #4
    i didnt get what you mean, sorry....

    the answer is 750 J, but i am confused with the formula...
  6. Sep 9, 2012 #5


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    Gold Member

    Ok, you have identified the two forces transferring energy away from the block, right?
    Compute the work done by those forces. (you should get negative values).
    Therefore, if x amount of joules is transferred away then the person must provide that same number of joules into the system.
  7. Sep 9, 2012 #6
    this is how i work on it:

    Total Force
    F = Wsin30 + fr
    = 200 (1/2) + 150
    = 250 N

    then, the path distance is 3m

    W = F(s)
    = 250 (3)
    = 750 J
  8. Sep 9, 2012 #7
    Im just confused with the sin30 thing, why is it that formula? thanks a lot.
  9. Sep 9, 2012 #8
    Lemme try

    now Frictional force =150N acting along the incline
    and weight of the box acting along the incline = 200 *sin(30)=200*1/2=100N(weight*sin 30 gives us its weight along the incline and weight*cos 30 gives us weight perpendicular to incline)
    So total force = 250 N along incline

    so the man is also exerting a force of 250 N

    now sin(30)=1.5/length of incline

    so length of incline = 1.5 *2 =3m

    so work done = 250*3 =750joules

    Hope u got it now
  10. Sep 9, 2012 #9

    thanks... i mean, why do we have to use sin30?
  11. Sep 9, 2012 #10
    Do u know any program that can be used to make drawings ??
  12. Sep 9, 2012 #11
    what kind of drawing? autocad, paint, corel draw, photoshop....? please accept my friend invitation.. thanks...
  13. Sep 9, 2012 #12
    Total work done=Conservation force x height + Non-conservative force x d.
    W=mgh + Ffd
    W=200x1.5 + 150 x 1.5/Sin30
    W=300 + 450
    W=750 J.
  14. Sep 9, 2012 #13

    now i can explain(sry for that messy thing)

    if u zoom u can see that mg is acting downward and can be resolved into 2 components

    1 along the incline and other perpendicular to it...(indicated by red)

    after resolving u can get mg in 2 parts

    here mgsin30 is acting along incline and mg cos 30 is acting along the perpendicular

    note that mgcos 30 is cancelled out by normal force exerted by block

    so mg sin 30 remains

    now friction also acts along incline

    so add forces along incline to get the force applied by man

    then proceed as said above
    Last edited: Sep 9, 2012
  15. Sep 9, 2012 #14
    sorry but i cant see the image..
  16. Sep 9, 2012 #15
    Download and zoom silly
  17. Sep 9, 2012 #16
    thanks a lot... its a great help... :)
  18. Sep 9, 2012 #17
    some other tips for such problems

    note that if friction is not given but coefficient of friction is given

    as u know frictional force = coefficient of friction*Normal force

    Here normal force would be mg cos 30

    But this is not need here as friction is already given

    hope that helped

    hoping to see u around....
  19. Sep 9, 2012 #18
    please add me as your friend because i know i will still have lots of questions,,, thans for the help...
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