Question

proof for law of conservation of energy for a body of vertically thrown upwards

Answer 1

case (a) : a ball is dropped from some height .
At A : let a ball of mass m is dropped from a height h . Here the total energy (T.E.) of the ball is the sum of kinetic energy (K.E.) and potential energy (P.E. ) .

potential energy = mgh
kinetic energy = 1/2 m(0)² =0
(T.E. ) = mgh + 0 = mgh




At B : Let the ball travel a distance of h¹ in time t during its fall . then the velocity of the ball after time t can be found by using equation of motion.

u=0 , a =g , s = h¹, v= v
using, v²-u² = 2as
v²-u² = 2gh¹
v² = 2gh¹
now, K.E. at B = ½mv² = ½m×2gh¹ = mgh¹
P.E. at B = mg (h-h¹)
(T.E.)b = K.E. +P.E. = mgh¹+mg (h-h¹) = mgh




therefore , total energy at A is equal to total energy at B . thus, energy remains conserved during a free fall .

If we analyse the fall in detail , we will find that as the ball falls down it loses its potential energy . That loss of potential energy appears in the form of gain in kinetic energy . therefore, we can also say that ,

loss in potential energy = gain in kinetic energy......




I HOPE IT HELPS...