Correct Answer - Option 3 : becomes 16 times of its initial KE
Concept:
Kinetic Energy (K):
The energy of the body in motion by virtue of motion is called kinetic energy.
\(K = \frac{1}{2}mv^2\)
m is the mass of the body, v is the speed of the body.
Calculation:
So, let the initial speed is u, and final speed is v
According to the question, v = 4u
Initial Kinetic Energy
\(K = \frac{1}{2}mu^2\)
Final Kinetic Energy
\(K' = \frac{1}{2}mv^2 = \frac{1}{2}m(4u)^2 = 16 (\frac{1}{2}mu^2)\)
⇒ \(K' = 16 K\)
So, the final kinetic energy is 16 times the initial KE.