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 Question

  1. Nov 1, 2005 #1
    Work Question..Anyone?!

    Im doing a problem and I found the work done by an applied force to be 401J.
    I then found the work done by gravity to be -330.8J
    Does this mean that i made an error..Arent (Wg + Wa) suppose to = 0
     
    Last edited: Nov 1, 2005
  2. jcsd
  3. Nov 1, 2005 #2
    There is a mass sitting on an incline and a force is applied so that it slides down at a constant speed.

    I found F=267N
    W=Fd....
    W=401J

    Wg=mgdcos(a)
    Wg=-330N

    Is d the total distance travelled..Or the y component of the distance travelled
     
  4. Nov 1, 2005 #3

    Pyrrhus

    User Avatar
    Homework Helper

    Please post the problem statement!
     
  5. Nov 1, 2005 #4
    Sure here is it

    A 45kg block of ice slides down a frictionless incline of 1.5m long and 0.9m. A worker pushes up against the ice parallel to the incline, so that the block slides down at a constant speed.
     
  6. Nov 1, 2005 #5
    a)find the magnitude of the workers force
    b)how much work is done on the block by the workers force
    c)" " the gravitational force on the block
    d)the normal force on the block from the surface on the incline
    e)the net force on the block
     
  7. Nov 1, 2005 #6

    Pyrrhus

    User Avatar
    Homework Helper

    Ok so show me your work, and/or where did you get stuck?
     
  8. Nov 1, 2005 #7
    drew the fbd's and came up with:

    -mgsin(theta) - F = -ma (a=0)
    -mgsin(theta) = F
    plug in numbers

    F = 267.5N
    Wf = Fd
    Wf = 267.5N(1.5m)
    Wf = 401J

    Wg = mgdcos(phi)
    Wg = 45kg(9.8m/s)(1.5m)(cos120)
    Wg = -330.8J

    This cant be correct because Wg +Wf = 0?
     
  9. Nov 1, 2005 #8

    Pyrrhus

    User Avatar
    Homework Helper

    i suppose the 1.5 and 0.9 are sides of the triangle right?
     
  10. Nov 1, 2005 #9
    1.5 = hypotenuse
    0.9 = height
     
  11. Nov 1, 2005 #10

    Pyrrhus

    User Avatar
    Homework Helper

    Ok i see the problem the mgcos(phi) component of gravity does not do work!, because it is perpendicular to the displacement vector, so the dot product will be 0!
     
  12. Nov 1, 2005 #11
    soo...how do i find the work done by gravity
     
  13. Nov 1, 2005 #12

    Pyrrhus

    User Avatar
    Homework Helper

    Gravity does work!, but only one component of gravity!, the one that is parallel to the displacement vector.
     
  14. Nov 1, 2005 #13
    but gravity is not perpendicular to the displacement vector..
    mgdcos(phi) makes it parallel to the displacement vector doesnt it?
     
  15. Nov 1, 2005 #14

    Pyrrhus

    User Avatar
    Homework Helper

    You have the gravity vector pointing down, and if you system of coordinates is directed at the angle the incline has, so the x axis is parallel to the hypotenuse, and the y axis is perpendicular to the hypotenuse, then you can decompose gravity in two components, one along the y axis (mgcos(angle)), and one along the x axis (mgsin(angle)), the displacement vector goes along the x axis.
     
  16. Nov 1, 2005 #15
    so to find the work done by gravity all i have to do is mgcos(30)
     
  17. Nov 1, 2005 #16
    is everything before this correct by the way?

    If so wouldnt Wg equal (-Wf)
     
  18. Nov 1, 2005 #17

    Pyrrhus

    User Avatar
    Homework Helper

    No, that's wrong i already explained why.

    Yes Wg IS equal -Wf.
     
  19. Nov 1, 2005 #18
    ok..but if i just mark this down i wont get any credit for it

    How could I prove that -Wf = Wg using my FBD's?
     
  20. Nov 1, 2005 #19

    Pyrrhus

    User Avatar
    Homework Helper

    I already say why :uhh:
    Try reading my #14 and #10 replies.
     
  21. Nov 2, 2005 #20

    verty

    User Avatar
    Homework Helper

    The forces are in equilibrium and gravity's working component is the equilibriant of the worker's force. Both of their magnitudes are given by mgsin(arctan(0.9/1.5)).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work Question
  1. Question on Work (Replies: 4)

  2. Work Question (Replies: 1)

  3. Work question (Replies: 1)

  4. Question on work (Replies: 3)

  5. Work question (Replies: 2)

Loading...