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## Work of an Object moving down Inclined Plane

1. The problem statement, all variables and given/known data
Experiment set up.
A 208g weight is attached via a pulley system to a block on an inclined plane. What is the work done by the suspended mass as the car is lowering at a constant velocity? And work done by gravity?

Distance weight moves down - 24.7cm
Incline is 30 degrees

2. Relevant equations
$F=ma$
$w=Fdcos(\theta)$

3. The attempt at a solution
So, in this case, work would be negative, right? Because the direction of the force from the suspended mass is going UP the incline, and the direction of moment is DOWN the incline?

The block is 474g

I'm not sure where to start. If I use the basic work formula and do
$\(-208*980)*24.7$?

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 Recognitions: Homework Help The thing losing energy is doing the work. The description of the experiment is incomplete - we are told about the suspended mass and then some car is introduced out of nowhere... but it looks like you are expected to use conservation of energy. Your descriptions of what you have tried are also incomplete so it is not clear what you have done. Try expressing your working symbolically - do all the algebra before you put numbers in.

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 Quote by Simon Bridge The thing losing energy is doing the work. The description of the experiment is incomplete - we are told about the suspended mass and then some car is introduced out of nowhere... but it looks like you are expected to use conservation of energy. Your descriptions of what you have tried are also incomplete so it is not clear what you have done. Try expressing your working symbolically - do all the algebra before you put numbers in.
Sorry, I wrote it really quickly.

The setup was an inclined plane with a pulley system which was set off the leg of the triangle. The pulley system was attached to the car. We had to set a weight on the pulley so that the car would go down the plane with a constant velocity.

I'm really not sure what to do. The car is moving down the car, therefore losing potential energy. If work can be defined as the difference in U.

So, if then ΔU= (474*980*12.5)-(474-980*0)= 5,806,500 erg?