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Work (done by gravity) - pushing up a ramp

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

Homework Statement



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.

Homework Equations



W=Fdcos(theta)

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
 

Answers and Replies

  • #2
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Does this maybe look right?

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

W = -1099 J
 
  • #3
Doc Al
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Does this maybe look right?

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

W = -1099 J
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.)
 
  • #4
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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
 
  • #5
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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 :(
 
  • #6
Doc Al
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Seems OK to me. Why do you think it's wrong?
 
  • #7
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cause stupid mastering physics says its wrong :(
 
  • #8
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Well at least i know i did it right. Thank you for checking it :)
 
  • #9
Doc Al
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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.
 
  • #10


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
 
  • #11
Doc Al
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W=(5)(9.80)(2.50)cos(30)
Why 30 degrees? (Realize that the angle between the force of gravity and the ramp is not 30 degrees.)
 
  • #12


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.
 
  • #13
Doc Al
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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.
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.
 

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