Work (done by gravity) - pushing up a ramp

• mybrohshi5
In summary: These two are not perpendicular, so you must use the correct angle to get the correct work.Thank you so much for your help! You're welcome. I'm happy to help, but I'm not sure you understand the issue I was trying to explain. Do you understand why the angle between the force of gravity and the displacement is 60 degrees? I'm not sure you do, and if not, I want to make sure you do, because otherwise you'll keep making the same mistake.I think I was just confused about the angle in general. I understand where you're coming from though. But I think I do get it now. I appreciate the help!
mybrohshi5
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.

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

Does this maybe look right?

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

W = -1099 J

mybrohshi5 said:
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.)

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

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 :(

Seems OK to me. Why do you think it's wrong?

cause stupid mastering physics says its wrong :(

Well at least i know i did it right. Thank you for checking it :)

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.

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

SParkinson13 said:
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.)

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.

SParkinson13 said:
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:
$$W = \vec{F}\cdot\vec{d} = Fd \cos\theta$$
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.

1. What is the work done by gravity when pushing an object up a ramp?

The work done by gravity when pushing an object up a ramp is equal to the force of gravity multiplied by the height of the ramp. This can be represented by the equation W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the ramp.

2. How does the angle of the ramp affect the work done by gravity?

The angle of the ramp affects the work done by gravity because it changes the height of the ramp. The higher the ramp, the more work gravity will do to push the object up it. This is because the higher the ramp, the more potential energy the object will have at the top.

3. Is the work done by gravity always positive when pushing up a ramp?

No, the work done by gravity can be either positive or negative when pushing up a ramp. If the ramp is inclined downwards, the work done by gravity will be negative as gravity will be pulling the object down the ramp. If the ramp is inclined upwards, the work done by gravity will be positive as gravity will be assisting in pushing the object up the ramp.

4. How does the mass of the object affect the work done by gravity when pushing up a ramp?

The mass of the object does not affect the work done by gravity when pushing up a ramp. This is because the formula for work (W = mgh) already takes into account the mass of the object. As long as the height of the ramp and the force of gravity remain constant, the work done by gravity will also remain constant.

5. Can the work done by gravity ever be greater than the work done by the applied force when pushing up a ramp?

Yes, the work done by gravity can be greater than the work done by the applied force when pushing up a ramp. This is possible if the ramp is very steep, as gravity will be doing more work to overcome the height of the ramp compared to the applied force. However, in most cases, the work done by the applied force will be greater than the work done by gravity.

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