Question

Prove the law of conservation of energy - i) When a body is thrown upwards ii) For a freely falling body​

Answer 1

ii.) For freely falling body

Law of conservation of energy states that:−

Energy can neither be created nor be destroyed.

Energy transforms from one form to another form and, potential energy+ kinetic energy= constant

Proof:−

At point A

K.E=0m/s

P.E=mgh

P.E=mgh

At point B,

K.E.,

2gh=V^2 −O^2

 V^2 =2gx

K.E=  1/2mV^2

K.E=  1/2 m.2gx

K.E=mgx-----------------------------(1)

P.E=mgh

P.E=m.g.(h−x)

P.E=mgh−mgx---------------------(2)

from(1) and (2)

K.E+P.E constant

mgx+mgh−mgx=mgh

mgh=mgh

At point C,  

P.E=0

2gh=v^2  −u^2

 ( initial velocity =0 m/s)

2gh=v^2

 K.E=  1/2.m.v

K.E=  1/2.m.2gh

K.E=mgh

Thus, in all the points the energy is same

i.) when the body thrown upwards:-

initial energy E = (1/2)mV^2...........................(1)

 

Let us consider a point at B which is at a height h above the ground level. Let v be the speed of body at B. Total energy of the body is given by

 

Energy at B = potential energy + kinetic energy = mgh+(1/2)mv^2  ...............(2)

 

But as per equation of motion v2 = V2-2gh    ............(3)

 

substituting for v^2 using (3) in (2), we get Energy at B = mgh+(1/2)m[V2-2gh] = (1/2)mV2 .............................(4)

 

Let C be the highest point the body can reach. Since the body comes to zero speed at this point, we have only potential energy

 

Energy at C = mgH ................................(5)

 

As per equation of motion we have, 0 = V 2-2gH ;  otherwise H = V 2/(2g) ..............(6)

 

By substituting for H using (6) in (5),  we get Energy at C = mgH = (1/2)mV2 ...............................(7)

 

Equation (7) shows that initial energy is fully converted into potential energy at the maximum height and both are same.

 

Let us assume during downward movement, at B, the falling body has the speed u m/s. Total energy EB at B during downward movement is given by

 

EB  = potential energy+kinetic energy = mgh+(1/2)mu2 .................(8)

 

as per equation of motion, u2 = 0+2g(H-h) ...................(9)

 

By substituting for u2 from (9) in (8), we habe EB = mgh+(1/2)m{ 2g(H-h) } = mgH .....................(10)

 

Finally when the body reaches A, it has only kinetic energy and its potential energy is zero

 

Energy EA at A at end of the downward movement is given by

 

EA = (1/2)mU2 ..................(11)

 

As per equation of motion U2 = 0+2gH ...............(12)

 

By substituting for U2 from (12) in (11), we get EA = (1/2)m{2gH} = (1/2)mgH ...................(13)