Question

If the velocity of a body is increased $4$ % then what is the approximate % change in its K.E.?

Answer 1

Let m= mass of the object and v= velocity of the object ,so kinetic energy of the object $E_{K.I}=\frac{1}{2}m{v}^2$------------(1)

Now as mentioned in the question velocity increased by 4%, thus kinetic energy of the object $E_{KF}=\frac{1}{2}m{\left(0.04v\right)}^2$-----(2)

Now comparing 1 and 2 we get,

Change in kinetic energy $\Delta E_K=\frac{1}{2}m(v^2-0.0016v^2)$

$\Delta E_K=49.92m$%