Potential Energy of .2kg Stone Thrown Upward

[superdave]

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

 

[Hootenanny]

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?

 

[superdave]

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?

 

[Doc Al]

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.)

 

[superdave]

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.

 

[Hootenanny]

Try to set up equations. What must the final energy equal?

 

[BobG]

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.

 

[Da-Force]

Go to my post

Go to my post on NRG.

https://www.physicsforums.com/showthread.php?t=114514"