Potential Energy of .2kg Stone Thrown Upward

Click For Summary

Homework Help Overview

The discussion revolves around the potential energy of a stone thrown vertically upward and the energy considerations of a roller coaster on a frictionless track. Participants are exploring concepts related to energy conservation and potential energy calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the potential energy at maximum height using kinematic equations but questions the validity of their approach. Some participants suggest considering energy conservation principles, while others inquire about the relationship between potential and kinetic energy at different points in the motion.

Discussion Status

Participants are actively engaging with the problem, with some offering guidance on energy conservation and questioning the original poster's calculations. There is a recognition of the need to clarify the distinction between height and potential energy, and the discussion is exploring multiple interpretations of the energy equations involved.

Contextual Notes

There is an emphasis on the conservation of energy in the context of both problems, and participants are noting the importance of understanding how mass factors into energy equations, particularly in the roller coaster scenario.

superdave
Messages
149
Reaction score
3
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
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
743
  • · Replies 21 ·
Replies
21
Views
2K
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
14K
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
8
Views
2K