Question

prove that the kinetic energy of a frely falling object on reaching the ground is nothing but the transformation of it's initial potential energy

give answer class 9th ls 2 work and energy​

Answer 1

Answer:

The object is at point A it's velocity is 0 and height is h

So,K.E is the work done on an object.

K.E = 1mv²

2

W = f×s

P.E = mgh.

1)K.E = 1mv²

2

Now v is the velocity which is 0 as the object is not moving.

Hence,K.E = 0

P.E = much

Total energy = kinetic energy + potential energy

= 0 + mgh

= mgh ............ (1)

2)The object comes down now it is at point B .

As it is moving its velocity is v.

K.E = 1 mv2

2

Now v²=u²+2as......Newton second law

Initial velocity (u)=0

Hence.v²=2as

Here acceleration is due to gravity g

V²=2gs

Displacement a is the covered distance= x

V²=2gh

Hence,K.E= 1 m2gx .........2 is cancelled

2

K.E = mgx

P.E = mgh here h is b-day

= .mg(h-x)

=mgh- mgx..bracket opens

Total energy = kinetic + potential energy

= mgx+mgh-mgx

= mgh.......... (2)

3) The object reaches ground at point C.

K.E= 1 mv²

2

v²= u²+2as (u=0)

= 2as

V=gh

a is gravity g

s is the total height= h

Hence,v²=2gh

K.E= 1 m2gh

2

K.E=mgh

P.E= mgh

Here h is 0

P.E =0

Total energy = mgh.........(3)

From 1,2 &3 it is proved.