Given,

Kinetic Energy (K_E) = 25 J

Velocity (v) = 5 ms^-1

We know,

Kinetic Energy (K_E) = $\dfrac{1}{2}$mv².

Substituting we get,

25 = $\dfrac{1}{2}$ x m x 5²

25 x 2 = 25 x m

m = $\dfrac{50}{25}$

m = 2 kg

When velocity is doubled :

v' = 10 ms^-1

m = 2 kg

Substituting we get,

K_E = $\dfrac{1}{2}$ x 2 x 10²

K.E. = 100 J

When velocity is increased three times, then

v'' = 15 ms^-1

m = 2 kg

Substituting we get,

K_E = $\dfrac{1}{2}$ x 2 x 15²

K.E. = 225 J

Hence, Kinetic Energy becomes 100 J when velocity is doubled and it becomes 225 J when velocity is increased three times.

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