Question

State Law of Conservation of Energy and express it in the form of an equation for a body of

mass m falling from a point A at height h, above the ground at (a) A, (b) B at a height from

ground, (c) C.

Answer 1

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Answer 2

Given:

Mass of body=m

Height of body at point A from ground=h

Solution:

Law of conservation of energy:

It states that energy can neither be create nor destroyed but it change from one form to another form.

At point A

P.E energy of body at height h from the ground=mgh

Initial velocity of body,u=0

K.E of body=0

Now, total energy of body=K.E+P.E=0+mgh=mgh

At point B

Height of body=h-x

P.E of body=mg(h-x)=mgh-mgx

$2gh=v^2-u^2$....(1)

Using the equation (1) of motion

$2gx=v^2$

Now, K.E of body

$K.E=\frac{1}[2}mv^2=\frac{1}{2}m(2gx)=mgx$

Total energy(T.E)=K.E+P.E

T.E=mgx+mgh-mgx=mgh

At point C

P.E of body=0

$2gh=v^2-u^2=v^2$

Now, K.E=$\frac{1}{2}m(2gh)=mgh$

T.E=mgh+0=mgh