Calculate the decrease in the kinetic energy of a moving body if it's velocity reduces to one third of the initial velocity
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If initial kinetic energy of the body is mv²/2 and its velocity decreases to one third.
Therefore the new kinetic energy, is
m(v/3)²/2 =(1/9)mv²/2
the decrease in kinetic energy = mv²/2-(1/9)mv²/2 = (8/9)mv²/2
percentage decrease =
{ {(8/9)mv²/2}/mv²/2}×100
8×100/9 =88.89%
Therefore the new kinetic energy, is
m(v/3)²/2 =(1/9)mv²/2
the decrease in kinetic energy = mv²/2-(1/9)mv²/2 = (8/9)mv²/2
percentage decrease =
{ {(8/9)mv²/2}/mv²/2}×100
8×100/9 =88.89%
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K.E.=1/2mv^2
From this formula of kinetic energy we get kinetic energy is directly proportional to the square of the velocity of the object
Now as velocity is reduced by 1/3 so kinetic energy will reduce by (1/3)^2=1/9
So New kinetic energy =1/18 mv^2
Decrease in kinetic energy =1/9 mv^2-1/18mv^2=3/18=1/6
From this formula of kinetic energy we get kinetic energy is directly proportional to the square of the velocity of the object
Now as velocity is reduced by 1/3 so kinetic energy will reduce by (1/3)^2=1/9
So New kinetic energy =1/18 mv^2
Decrease in kinetic energy =1/9 mv^2-1/18mv^2=3/18=1/6
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