1. Limited time only! Sign up for a free 30min personal 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 (done by gravity) - pushing up a ramp

  1. Feb 28, 2010 #1
    Work (done by gravity) --- pushing up a ramp

    1. The problem statement, all variables and given/known data

    A physics professor is pushed up a ramp inclined upward at an angle 28.0^\circ above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 81.0 kg. He is pushed a distance 2.95 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor's speed at the bottom of the ramp is 2.15 m/s.

    2. Relevant equations

    W=Fdcos(theta)

    3. The attempt at a solution

    I know that if the force of gravity is perpendicular to the force applied then the work done by gravity is zero.

    Since the force applied to the professor and chair is horizontal and the force due to gravity on the prof. and chair is vertical i thought the answer would be 0.

    This is wrong though. Any suggestions?

    Thank you
     
  2. jcsd
  3. Feb 28, 2010 #2
    Re: Work (done by gravity) --- pushing up a ramp

    Does this maybe look right?

    W = 81(-9.8)(2.95*sin(28))

    W = -1099 J
     
  4. Feb 28, 2010 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Work (done by gravity) --- pushing up a ramp

    That's the work done by gravity.

    What are you asked to find in the problem? (I suspect they want the final speed of the professor at the top of the ramp.)
     
  5. Feb 28, 2010 #4
    Re: Work (done by gravity) --- pushing up a ramp

    Ok that was right. Now i really need help with this next part cause i cannot figure out what i am doing wrong and i only have 1 attempt left to answer it :(

    Find the speed of the professor-chair unit at the top of the ramp

    Fd=1/2mvf2 - 1/2mvi2

    I found the sum of the forces in the x direction for the F to be used in this equation above.

    F = 600cos(28) - 81(9.8)(sin28)

    F = 158

    158(2.95) + 1/2(81)(2.152) = 1/2(81)(vf2)

    vf = 4.02 m/s

    I cannot find where i went wrong with this one :(

    Any suggestions?

    Thank you
     
  6. Feb 28, 2010 #5
    Re: Work (done by gravity) --- pushing up a ramp

    Sorry....I was asked both the work done by gravity and yes now the final speed at the end of the ramp, which i found above, but its wrong :(
     
  7. Feb 28, 2010 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Work (done by gravity) --- pushing up a ramp

    Seems OK to me. Why do you think it's wrong?
     
  8. Feb 28, 2010 #7
    Re: Work (done by gravity) --- pushing up a ramp

    cause stupid mastering physics says its wrong :(
     
  9. Feb 28, 2010 #8
    Re: Work (done by gravity) --- pushing up a ramp

    Well at least i know i did it right. Thank you for checking it :)
     
  10. Feb 28, 2010 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Work (done by gravity) --- pushing up a ramp

    Mastering Physics can be fussy about significant figures sometimes. It can't hurt to redo your calculations with greater accuracy, only rounding off at the last step.
     
  11. Jun 15, 2010 #10
    Re: Work (done by gravity) --- pushing up a ramp

    I had a question simular to the one posted above. I solved it the way shown and it's not matching the answer in my book.
    Question:
    Starting from rest, a 5.00-kg block slides 2.50m down a rouch 30.0 degree incline. The coefficient of kinetic friction between the block and the incline is 0.436. Determine the work done by the force of gravity.

    I thought you could just do

    W=(5)(9.80)(2.50)cos(30)
    W= 106J

    but the answer is 61.3J
     
  12. Jun 15, 2010 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Work (done by gravity) --- pushing up a ramp

    Why 30 degrees? (Realize that the angle between the force of gravity and the ramp is not 30 degrees.)
     
  13. Jun 15, 2010 #12
    Re: Work (done by gravity) --- pushing up a ramp

    Oh, I was supposed to use 60 degrees! Do I always have to use the angle between the force of gravity and the ramp? Or is that only when solving for force of gravity? I've honestly never been told by my proff. to do this. But he doesn't teach us very much.
     
  14. Jun 16, 2010 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Work (done by gravity) --- pushing up a ramp

    Whenever you are computing the work done by a force acting through a displacement, like this:
    [tex]W = \vec{F}\cdot\vec{d} = Fd \cos\theta[/tex]
    The angle you use is the angle between the two vectors. In this case that angle is 60 degrees. In this problem the force is gravity, which acts straight down, and the displacement is along the ramp.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work (done by gravity) - pushing up a ramp
  1. Work Problem: Up a ramp (Replies: 10)

Loading...