Question

Show that the sum of potential and kinetic energy of a freely

falling body is always constant.​

Answer 1

Answer:

Now, P=mgh and K=mv²/2

so, here, when the object will be on a higher position , then

Let's consider this as A

THEREFORE

over here , Kinetic energy=0 as the object is static and potential energy is full, complete and highest so mgh.

now, sum at this position A =0+mgh=MGH(constant)

If the object will cover half distance with the force of graviration as 9.8N

Then, let's regard it as position B

therefore, p=mgh/2 and

For calculating kinetic energy

v²=u²+2as

so, here , u=0

hence, v²=2as

so a=g and s=h/2

therefore v²=2×g×h/2

so, v²=gh

now, kinetic energy=mv²/2= mgh/2

so the sum of kinetic energy and potential energy =mgh/2+mgh/2=MGH(constant)

If the object will fall on the ground then kinetic energy would be highest and let's regard it as position C

therefore, mv²/2 =(m×g×h×2)/2

so, mgh

and, potential energy=0

now ,

kinetic energy + potential