# Proof for potential energy + kinetic energy is a constant when there is no energy loss

Recently I learned that if there is no energy loss the sum of potential energy and the kinetic energy is a constant. So I tried to prove that assuming that using the below diagram

When t=0 I got,
Kinetic energy = 0 (The object is still)
Potential energy = mgh
Potential energy+Kinetic energy=mgh

When t=1 I got,
Kinetic energy=(1/2)mg^2(when t=1 the velocity is g because gravitational acc. is g ms-1)
Potential energy=mg(h-g)=mgh-mg^2(When t=1 the object has travelled g distance)
Potential energy + Kinetic energy=mgh-(1/2)mg^2

According to the theory what I've learned the sum of E(p)+E(k) is equal regardless of the time so then It should be
mgh=mgh-(1/2)mg^2

But mgh is not equal to mgh-(1/2)mg^2 which means I have gone wrong somewhere I even tried to get factors and simplify some more but the answer was the same and also I repeated the same to t=2 and t=3 that also gives it is not constant. Could anyone in this community help me to prove this? Thank you.

Edit: I forgot to u=(1/2)at^2 fortunately @chris97ong showed me it in the comments (Thanks chris)

Now the proof is complete and I got mgh as the sum on any value for t

Thanks to everyone who afford their time to this.

• When $t=1$, distance covered is $\frac{1}{2} g t^2$, which is $g/2$, not $g$. Commented Oct 28, 2021 at 7:04