Physics, asked by nikitapal2167, 4 days ago

A pebble is thrown up. It goes to a height and then come back on the ground. State the different changes in form of energy during its motion.


Answered by abhi178
Assume a pebble is thrown up with velocity u . It goes to a height h and then comes back on the ground , before striking velocity of pebble is - u m/s

Initially, at lowest position ,
pebble has only kinetic energy
e.g , K.E = 1/2 mu² , here m is the mass of pebble.
Totol energy = K.E + P.E = 1/2mv² + 0 = 1/2 mu²
At heighest position ,
pebble has only potential energy
e.g., P.E = mgh
Total energy = K.E + P.E = 0 + mgh = mgh

According to conservation of energy ,
Total energy of system of particle is conserved.
so, total energy at lowest position = total energy at heighest position
e.g., 1/2mu²= mgh , u = √(2gh) m/s

Now, at middle position,
height = h/2
velocity = √(2gh/2) = √(gh)
Now, total energy = K.E + P.E = 1/2m(√(gh))² +mgh/2
= mgh/2 + mgh/2 = mgh = total energy at heighest position

It is clear that total energy at any position of motion is always same.
Hence, it shows body shows conservation of energy .
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