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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 diagramenter image description here

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.

Sorry if the question is unclear :) (If it is really unclear please leave a comment)

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.

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  • $\begingroup$ When $t=1$, distance covered is $\frac{1}{2} g t^2$, which is $g/2$, not $g$. $\endgroup$
    – chris97ong
    Commented Oct 28, 2021 at 7:04

1 Answer 1

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This line

Potential energy=mg(h-g)=mgh-mg^2(When t=1 the object has travelled g distance)

Should be

"Potential energy=mg(h-0.5g)=mgh-0.5mg^2(When t=1 the object has travelled 0.5g distance)"

It's true that the final speed after 1 second is g, but since it started from rest, the average speed is 0.5g m/s over the first second, so it falls a distance 0.5g.

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  • $\begingroup$ Thanks @John Hunter $\endgroup$ Commented Oct 28, 2021 at 10:19

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