Help Deriving the equation Ek=1/2mv^2

1. Jan 12, 2006

skull

Ok these are the equations im allowed to use.
Ep=mg(delta)h
w=fd
v_av=(delta)d/(delta)t
(delta)d=v1(delta)t^2 + 1/2a(delta)t^2
V2=V1^2 + 2a(delta)d
w=work done(j)
f=force(newtons)
d=distance(m)
v_av=average velocity
t=time(secs)
v2=final velocity
v1=initial velocity
a=acceleration

Ok, i havent done anything so far, because I dont know where to start. Im not asking anyone to solve this for me, i'm just asking for some helpful clues
thanks

2. Jan 12, 2006

daniel_i_l

You could use energy conservation with the potential energy and use kinimatics to find the final speed after a change in height, using that you should get the same answer that you would have gotten with KE so you would be able to derive it. (I hope that was clear)
But thats probably cheating, are you alowed to use cal ? If so than just sum up all of the work done over an interval.

3. Jan 12, 2006

skull

Ok, first question, where does the Ek come from. It is not in any other equation, doesnt it have to be in order for me to derive i?

4. Jan 12, 2006

skull

anyone know?

5. Jan 12, 2006

G01

Using Newton's Law's you can derive the conservation of mechanical energy. In essence you can derive the expression, 1/2mv^2 + mgh = 0. You then define 1/2mv^2 as kinetic energy. Is this what your trying to do?

6. Jan 12, 2006

skull

i dont fully understand that part

7. Jan 12, 2006

lightgrav

Using Calculus:
Write down Newton #2 (show a time derivitive)
Most situations have Force functions that depend on location (x) rather than time.
So multiply Newton#2 by dx, which can switch places with the dv.
Now find the antiderivitive of each side.

Without Calculus:
v^2 = v^2 + 2ax ...
multiply by mass, then replace Fx by Work

Last edited: Jan 12, 2006
8. Jan 12, 2006

robphy

Ek arises as an interesting quantity from the "Work-[Kinetic]Energy Theorem". Start with Newton-II for the case of a constant net force F_net doing work over a displacement d. Use your kinematic equations for constant acceleration to reveal this interesting quantity.

Before you start, you should really check the correctness of the equations that you were allowed to use... although the errors that appear may just be typos.

9. Jan 13, 2006

daniel_i_l

I was thinking of using the fifth equation (V^2 = ...) and the first one. Look what happens if you divide the fifth equation by 2 and see how the final speed is related to initial hight. Multiplying by m should easily let you get to mV^2 / 2