Question

The kinetic energy of an object of mass, m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Answer 1

Kinetic energy, k = \(\frac{1}{2}\) mv².
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 k₁ = 25J
Initial velocity V₂ = 5 ms^-1
New kinetic Energy K₂ = ?
New velocity v₂ = 3v₁ = 3 × 5 = 10 ms^-1

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