Gravitational Potential energy on an incline

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SUMMARY

The discussion clarifies that gravitational potential energy (PE) is calculated using the formula PE = mgh, where g is the acceleration due to gravity, consistently valued at 9.8 m/s² regardless of the incline angle. The potential energy solely depends on the height difference (h2 - h1) between the starting and ending points, not on the angle of the ramp. Therefore, scenarios involving different ramp angles do not affect the value of g in the potential energy calculation.

PREREQUISITES
  • Understanding of gravitational potential energy (PE) and its formula PE = mgh
  • Basic knowledge of physics concepts related to motion and forces
  • Familiarity with the concept of height in relation to gravitational force
  • Knowledge of the MCAT exam structure and content
NEXT STEPS
  • Review the principles of gravitational potential energy in different contexts
  • Study the relationship between potential energy and kinetic energy in physics
  • Explore the effects of different angles on forces acting on objects on inclines
  • Practice solving problems related to energy conservation in physics
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Students preparing for the MCAT, physics learners, and educators seeking to reinforce concepts of gravitational potential energy and its applications in various scenarios.

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Hello all, I'm new to the forum here and desperately need some help in some basic physics concepts that I need to have a firm grasp on for the MCAT.

My question is about PE on an inclined plane. For example, there are 3 possible scenarios (the height and mass remains the same for each scenario). Scenario 1 has a ramp angled at 30, scenario 2 has a ramp angled at 60, and the last scenario has no ramp at all- the person must lift it from the ground to the desired height.

What I am asking is that when PE (or U)= mgh, does g remain the same at 9.8m/s^2 or is it affected by the angle of the ramp, ie gsin30? And could you please explain why?

I know when you solve for acceleration in an unrelated problem to PE you do, but I'm not sure about for just when solving for PE.

Thanks!
 
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potential energy only depends on the height, mgh. whatever the difference in height at the start and end, mg(h2-h1) will be the change in potential energy, either becoming kinetic (rolling down) or becoming potential (going up ramp). the potential has nothing to do with the ramp, only the change in height. g never changes
 

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