K.E of a particle depends on mass and velocity.show that the K.E of the particle is 1/2mv^2
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we know that---
v^2-u^2=2as
=>s=(v^2-u^2)÷2a. ................(i)
Again,W=Fs
=>K.E=ma×v^2-u^2÷2a....[from (i)]
=1/2m(v^2-u^2)
And if u=0,then--
K.E=1/2mv^2
v^2-u^2=2as
=>s=(v^2-u^2)÷2a. ................(i)
Again,W=Fs
=>K.E=ma×v^2-u^2÷2a....[from (i)]
=1/2m(v^2-u^2)
And if u=0,then--
K.E=1/2mv^2
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