Kinetic energy, k = \(\frac{1}{2}\) mv^{2}.

Where m = mass of the object

V = velocity of the object.

Here mass (m) is the same in both cases (∵ object is same)

\(\frac{k_1}{k_2}=(\frac{v_1}{v_2})^2\)

Initial kinetic energy k_{1} = 25J

Initial velocity V_{2} = 5 ms^{-1}

New kinetic Energy K_{2} = ?

New velocity v_{2} = 3v_{1} = 3 × 5 = 10 ms^{-1}

∴ When velocity is increased three times its K.E. is 225 J.