Question

CHAPTER: WORK,POWER,ENERGY
TOPIC: LAW OF CONSERVATION OF ENERGY
Q. Prove the law of conservation of energy is not violated in the figure sharing below, at a height of 10m, 5m and just before touching the surface of the water body .
222 Hints:Tousetheformulav-u=2gh .[g=10m/s]
Mass of the diver is 75kg.

Answer 1

(The law of conservation of energy states that, energy cannot be destroyed or created. It can be transferred from one form to the other and it remains constant before and after transformation.)

Here we have to show that the total mechanical energy is constant.

Or, PE + KE = constant

Case 1:- (10 m)

We know, PE = mgh

⇒ PE = (75 kg)(10 m/s²)(10 m)

⇒ PE = 7500 J

And, KE = ½ mv²

Since she started from rest, ⇒ v = 0.

∴ KE = ½ m × 0² = 0

Or, total mechanical energy = 7500 J + 0 = 7500 J

Case 2:- (5 m)

Here, PE = mgh = (75 kg)(10 m/s²)(5 m) = 3750 J

& KE = ½ mv²

∵ v² = 2gh ...(A)

∴ KE = ½ m × 2gh

⇒ KE = ½ × 75 kg × 2 × 10 m/s² × 5 m

⇒ KE = 37.5 kg × 100 m/s² × m

⇒ KE = 3750 J

∴ Total mechanical energy = 3750 J + 3750 J = 7500 J. Or, the energy here is conserved.

Case 3:- (0 m)

Here, PE = mgh = mg × 0 = 0 J

And, KE = ½ mv² = 37.5 kg × v²

⇒ KE = 37.5 kg × 2gh (from A)

⇒ KE = 75 kg × 10 m/s² × 10 m

(Displacement = 10 m)

⇒ KE = 7500 J

∴ Total mechanical energy = 0 J + 7500 J = 7500 J.

So, the law of conservation of energy is not violated in this case.