Physics 11 Force and potential energy

AI Thread Summary
A constant net force of 410N is applied to a stone weighing 32N, moving it upward over a distance of 2.0m before release. The potential energy equation Ep=mgh is discussed, with attempts to calculate the height the stone will rise using various methods, including impulse and the Work-Energy theorem. The Work-Energy theorem is emphasized as a suitable approach, where the work done by the force is equated to the potential energy at maximum height. After calculations, a height of 26m is suggested as the answer. The discussion highlights collaborative problem-solving in physics, with participants offering insights and support.
Iceclover
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Homework Statement



A constant net force of 410N up is applied to a stone that weighs 32N. The upward force is applied through a distance of 2.0m andthe stone is then released. TO what height from the point of release will the stone rise?

Homework Equations



Ep=mgh

The Attempt at a Solution



Ep=mgh
410=32(h)
h=13m
 
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I'm not sure where you're going with your solution.
Personally, I'd apply impulse again to get the stone's final velocity at the time of release, and then treat it as one-dimensional motion with constant acceleration.
 
i can't use impulse can I? i don't have a time?
 
Picky, picky...:blushing:

OK, how about the Work-Energy theorem to find the velocity at release?
 
Ek=1/2mv^2? and which force do i use
 
Iceclover said:
Ek=1/2mv^2?

Ah, no. Can you state the Work-Energy Theorem? (If not, can you look it up in your textbook or notes?)
 
An easier way is to use the Work-Energy Theorem, as stated above. Calculate the work done by the force on the stone for the 2m that it is applied (recall, work done = force * distance). When the stone reaches its maximum height, all the energy is the potential energy, so you can equate the potential energy to the work done, and solve for h.
 
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Work done on object is equal to the change in its kinetic energy
 
OK, so can you apply it? (I know I can...)
 
  • #10
what do i do once I've found velocity
 
  • #11
Use a basic kinematics equation for motion with constant acceleration.
 
  • #12
the answer i got was 26m
 
  • #13
Oddly enough, so do I.
 
  • #14
ok good, thank you very much
 
  • #15
do you have time for another question?
 
  • #16
Not really... but I'm in the middle of studying for my physics final, so this helps me, too. Look, If I don't answer, someone else will. (That's why this forum rocks!)
 
  • #17
ok sounds good, i think I am going to make a new topic though
 
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