Gravitational Potential Energy on an Incline

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Gravitational Potential Energy (GPE) is typically calculated using the formula U=mgh, where m is mass, g is the acceleration due to gravity, and h is the height above a reference point. On an incline, the height (h) must be measured vertically from the reference point, not along the incline. Therefore, while the formula remains the same, the effective height (h) changes based on the angle of the incline. For an incline at 52 degrees, the vertical height can be calculated using trigonometric functions. Understanding how to determine the correct height is crucial for accurately calculating GPE on an incline.
Travis Enigma
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Homework Statement
What is the Gravitational Potential Energy on an Incline? Is it still mgh?
Relevant Equations
U=mgh
Hi,

When regarding Gravitational Potential Energy, I know the formula is U=mgh. However, when the object is on an incline (say at an angle of 52 degrees) would it still be mgh or something else? (This isn't homework I simply was just curious).
 
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Travis Enigma said:
Homework Statement:: What is the Gravitational Potential Energy on an Incline? Is it still mgh?
Relevant Equations:: U=mgh

Hi,

When regarding Gravitational Potential Energy, I know the formula is U=mgh. However, when the object is on an incline (say at an angle of 52 degrees) would it still be mgh or something else? (This isn't homework I simply was just curious).
Well, whar do the letters m, g, h mean?
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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