A ball is dropped vertically from a height of 12m from rest. After striking the ground, it bounces to a height of 9m. What fraction of Kinetic Energy it lose on striking the ground?
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Let v = velocity of the ball just before it strikes the ground
We know
v2 = 2gh
Multiplying both sides by m
mv2 = 2mgh
=> ½mv2 = mgh
Now loss in KE = ½mv12 – ½mv22
=> ½mv12 – ½mv22 = mgh1 – mgh2
Loss in KE = Loss in PE
Potential energy of the ball at height 12 m is
PE1 = mg12
Potential energy of the ball at height 9 m is
PE2 = mg9
Loss in energy = PE1 – PE2 = mg(12 - 9)
= 3mg
Fraction of KE lost = 3mg / 12mg
= ¼ or 25%
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Let v = velocity of the ball just before it strikes the ground
We know
v2 = 2gh
Multiplying both sides by m
mv2 = 2mgh
=> ½mv2 = mgh
Now loss in KE = ½mv12 – ½mv22
=> ½mv12 – ½mv22 = mgh1 – mgh2
Loss in KE = Loss in PE
Potential energy of the ball at height 12 m is
PE1 = mg12
Potential energy of the ball at height 9 m is
PE2 = mg9
Loss in energy = PE1 – PE2 = mg(12 - 9)
= 3mg
Fraction of KE lost = 3mg / 12mg
= ¼ or 25%
plz mark as brainliest
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