Work Done by Incline Plane & Gravity on Block?

AI Thread Summary
In a discussion about a block sliding down a frictionless inclined plane, it was clarified that the work done by the incline on the block is zero, as the normal force acts perpendicular to the block's movement. Conversely, the work done by gravity is considered positive since gravity is the force that propels the block downward along the incline. The relationship between gravitational potential energy and kinetic energy was also highlighted, emphasizing that as the block descends, it loses potential energy while gaining kinetic energy. The equations for kinetic energy (KE = 1/2 mv²) and potential energy (PE = mgh) were referenced to support these points. Overall, the discussion confirmed the initial assumptions regarding work done by both the incline and gravity.
TheNotoriousWM
Messages
2
Reaction score
0
Say we have a friction-less inclined plane in the shape of a right triangle, sloping downward to the right. The angle above the horizontal axis is inconsequential for these specific purposes aside from the fact that we know it is inclined above said axis.

We then have a block sliding down (to the right) the plane. My question is about the work done on the block by the incline plane and gravity.

Is the work done by the inclined place on the block considered to be positive, negative, or zero? I think it is zero as what would be considered the normal force has a direction perpendicular to the movement of the block.

I am also hung up on the same question regarding the work done by gravity. It would make sense to think the work by gravity would be positive as that is what moves the block down the inclined plane without outside forces having acted upon it. I just wanted to verify these if anyone could possibly clear things up a bit.

Thanks in advance.

12540n8.jpg
 
Physics news on Phys.org
If you lift an object, you are giving it gravitational potential energy.

So, to put it at the top of that incline, it will have gained potential energy.

When it is released, it is losing potential energy, but gaining gravitational kinetic energy.

KE = PE

==================

When it's frictionless, it's best to think of these problems with the equations

KE = \frac{1}{2}mv^{2}

PE = mgh
 
TheNotoriousWM said:
Say we have a friction-less inclined plane in the shape of a right triangle, sloping downward to the right. The angle above the horizontal axis is inconsequential for these specific purposes aside from the fact that we know it is inclined above said axis.

We then have a block sliding down (to the right) the plane. My question is about the work done on the block by the incline plane and gravity.

Is the work done by the inclined place on the block considered to be positive, negative, or zero? I think it is zero as what would be considered the normal force has a direction perpendicular to the movement of the block.

I am also hung up on the same question regarding the work done by gravity. It would make sense to think the work by gravity would be positive as that is what moves the block down the inclined plane without outside forces having acted upon it. I just wanted to verify these if anyone could possibly clear things up a bit.

Thanks in advance.

12540n8.jpg
Hello TheNotoriousWM. Welcome to PF !

You are correct on both accounts.
 
Thanks for the help all, never hurts to check.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top