Potential Energy of .2kg Stone Thrown Upward

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SUMMARY

The discussion focuses on calculating the potential energy of a 0.2 kg stone thrown vertically upward with an initial velocity of 7.5 m/s from a height of 1.2 m. The incorrect application of the kinematic equation led to confusion, as the user mistakenly calculated height instead of potential energy. The correct approach involves using the conservation of energy principle, where the total mechanical energy at the start equals the potential energy at maximum height. The user is advised to consider energy conservation for further calculations, particularly in scenarios involving roller coasters on frictionless tracks.

PREREQUISITES
  • Understanding of kinematic equations, specifically v^2 = v0^2 + 2a(y - y0)
  • Knowledge of the conservation of energy principle in physics
  • Familiarity with potential energy calculations, U = mgh
  • Basic understanding of kinetic energy, K = 0.5mv^2
NEXT STEPS
  • Study the conservation of mechanical energy in closed systems
  • Learn how to derive potential energy from height and mass
  • Explore the relationship between potential and kinetic energy in roller coaster dynamics
  • Investigate the impact of friction on energy conservation in real-world scenarios
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics, energy conservation, and kinematic equations. This discussion is also beneficial for anyone studying roller coaster physics and energy transformations.

superdave
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Alright, another one I am messing up somehow

.2kg stone thrown vertically upward with initial vof 7.5 starting at 1.2 m.

Find potential energy at max height.

I use

v^2=v0^2+2a(y-y0)
solving for y:
y=-v0^2/2a+y0 = 7.5^2/(2*9.8) + 1.2 = 6.7 J
the answer is wrong
 
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This one may be easier if you consider energy. Remember that energy is conserved. How much energy do you have at the start? Therefore, how much energy must you have at max height?
 
Hrm. So I find U+K for the start. And then it'll all be K at the max height?

I'll do that, but can you tell me what's wrong with the way I did it?
 
superdave said:
solving for y:
y=-v0^2/2a+y0 = 7.5^2/(2*9.8) + 1.2 = 6.7 J
the answer is wrong
The height is not the potential energy. Use it to find the potential energy. (If you checked your units, you'd be less likely to make this kind of mistake.)
 
Okay, new question

Rollar coaster, frictionless track.
Speed at A (5.0 M, top of a hill)=5.0 m/s, what is the speed at B (bottom of hill).

I don't know how to solve this using the energy equations, because we don't know the mass.
 
Try to set up equations. What must the final energy equal?
 
superdave said:
Okay, new question

Rollar coaster, frictionless track.
Speed at A (5.0 M, top of a hill)=5.0 m/s, what is the speed at B (bottom of hill).

I don't know how to solve this using the energy equations, because we don't know the mass.
You do know that the sum of potential energy and kinetic energy at the top equals the kinetic energy at the bottom. If you set up the equations, you'll notice that the mass cancels out of both sides of the equation.
 

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