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# Little question

A mass m drops of from a height h (vertically). Calculate the speed at the bottom (1) and the kinetic energy (also at the bottom). (2)

Now this question is haunting me... Because normally I get exercises that are around one or two pages to solve and now I got this one, which I solved in two lines, so I'm really doubting my answer. I solved it as such:

Potential energy (Gravitational energy) = m*g*h = 10*m*h Because the potential energy changes in Kinetic energy, we can substitute those two...
Kinetic energy = (m*v^2)*0.5 = 10*m*h (2)
v = V(20*h) (1)

Now I'm really asking myself if it's correct, so I'm hoping someone can bring some light to this (easy) question.

_AVG_
Yup, I would say that your answer is correct : v^2 = 2gh (in accordance with equating kinetic energy and potential energy as well as solving Newtonian equations for uniform acceleration)

But this is based on a few assumptions:
- h is so small that g remains constant
- air resistance is neglected (or h is so small that air resistance has a negligible effect)
ofertas
particle acceleration
Bikerman
Looks good to me (with the provisos already given by _AVG_).

PS - I'm assuming, of course, that you have used V to mean the square root radix, and it would be better to express the answer in g rather than round it to 10 methinks
chatrack
Hi,

You are right

Potential energy at the height h = Kinetic energy at the last moment (instant) of its fall
Dennise
Yup!

Energy conservation is one of the coolest and fastest way to approach physics problems. It can save gobs of time in finding a problem's solution.
tazone
only true if you dont take the speed relativity in consideration (relative to light speed, would make a small number change in the equation)

also its only true if there is no friction, no heat loss or other forms of energy going into different states than the ones in your formula