# 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.

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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 .

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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