Question

prove that the sum of kinetic energy and potential energy of a freely falling body remains always constant .

Answer 1

Answer:

At A,velocity =0=u

KE=  

2

1

​  

mv  

2

=0

PE=mgh,TE=PE+KE=mgh

At B,velocity v:

v  

2

=u  

2

+2as

v  

2

=2gx

KE=  

2

1

​  

mv  

2

=mgx

PE=mg(h−x)

Total energy TE=PE+KE

=mg(h−x)+mgx=mgh

At C,velocity v:

v  

2

=u  

2

+2as

v  

2

=2gh

KE=  

2

1

​  

mv  

2

=mgh

PE=0

TE=PE+KE=mgh