Calculate kinetic energy if velocity is reduces by 1/3rd
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Let KE(1) =m/2 ×[v(1)]^(2),where m is mass and v is initial velocity. we know that V(2)=V(1)/3.
So, KE(2)=m/2×V(2)^(2)
=m/2×[V(1)/3]^(2)
=m/2×V(1)^(2)/9
=[m×V(1)]÷18.
KE(2)=KE(1)/9
i.e. KE(2) is one by ninth of KE(1)
So, KE(2)=m/2×V(2)^(2)
=m/2×[V(1)/3]^(2)
=m/2×V(1)^(2)/9
=[m×V(1)]÷18.
KE(2)=KE(1)/9
i.e. KE(2) is one by ninth of KE(1)
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