# Forces and Work on an Incline...

1. Oct 29, 2015

### Tim Wellens

1. The problem statement, all variables and given/known data
The question is about a block on an incline at an angle Θ above the horizontal. The block will begin from the resting position and travel from point A to B down the incline. Neglect friction.

1) As the block travels from point A to B, will the magnitude of the net force exerted on the block increase, decrease, or stay constant?
2) For each force identified in your free body diagram, state whether the work done on the block is positive, negative, or zero

2. Relevant equations
W=FdcosΘ

3. The attempt at a solution
1) I think the net force on the block, as it travels from point A to B, would remain constant because the variables that make up the forces stay constant. I think there will be a constant net force exerted on the block until the block comes to a rest. I think this net force would be ΣF=mgsinΘ because ΣFy=n-mgcosΘ=0 due to the lack of movement in the y direction, which makes the net force ΣF=mgsinΘ. This net force I think would stay the same?

2) The forces that I listed that act on the block are the normal force and the weight force. The normal force would have positive work done, i think, because if we use W=FdcosΘ.. then the force is positive because it would be mgcosΘ.. The force and distance will be positive, as is the angle. Or is the distance 0 because the block isn't traveling at all in the y direction?

The weight force i'm confused on because you can break the weight force up into components, the y component is negative and the x is positive. So, I'm not sure if the work done by the weight force would be negative, positive or zero..

2. Oct 29, 2015

### Geofleur

If the angle between the force and the displacement vector is 90 degrees, no work will be done because the cosine of 90 degrees is zero. Draw the displacement vector and the forces in your problem. Are any of them perpendicular to it? Similarly, if the angle between the force and the displacement vector is less than 90 degrees, the work done will be positive. Knowing these things now, would you like to modify any of your answers?

3. Oct 29, 2015

### Tim Wellens

Oh, ok. So, the normal force will be zero because it's force is perpendicular to the displacement vector, and the weight would be positive because it's force vector makes an angle of less than 90 degrees with the displacement vector. So, we always use the weight without breaking it into components for figuring out the work? To figure out the net work I would add the work done by each of the forces, so in this case we only have normal force and weight force.. So the work would be positive because we only have zero work plus positive work, correct?

4. Oct 29, 2015

### haruspex

Yes, that's about right. I assume you meant the work done by the normal force is zero.
Whether you break the force into components is up to you. Sometimes it's easier that way. Should get the same result.
In your post #1, you seem to shift between Y being vertical and Y being normal to the plane.

5. Oct 30, 2015

### Tim Wellens

I think Y is vertical to the plane and the forces in the y direction equal 0, while the forces in the x direction equal mgsinΘ.So, the net force is mgsinΘ. Is it correct that the net force would remain constant as the block travels from point A to B because the mass, gravity, and angle stay constant as the block travels? I can't think of any reason why it would decrease or increase.

6. Oct 30, 2015

### haruspex

You mean normal to the plane. Vertical is normal to the horizontal.
And it is whichever you choose it to be. It helps readers if you mention which you are choosing.
Yes.