Potential energy on incline

[brood910]

According to TBR, on an incline, PE = mghsin(theta). I thought PE is only dependent on vertical components? So isnt it still PE = mgh?

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[eradicator]

According to TBR, on an incline, PE = mghsin(theta). I thought PE is only dependent on vertical components? So isnt it still PE = mgh?

Yes it should still be mgh, h being the height of the incline.
Where did u find that in TBR?

 

[navho77]

I agree, PE should still be mgh. mgsin(theta) is significant because it is the force that's doing the work, ie converting the PE into KE (and heat if there is friction). mgsin(theta) is acting through the distance that corresponds to the length of the incline, which can be written as h/sin(theta). So, the the total amount of work done is mgsin(theta) x h/sin(theta)=mgh. From energy conservation, the total amount of work done is maximally equal to the original energy of the system, so the original energy must have been mgh.