Question

how does the kinetic energy change if mass is increased 2 times and velocity is increased 3 times on the same object​

Answer 1

Explanation:

let velocity =v,mass=m

therefore inital Kinetic energy(KE)=1/2mv^2

now mass is made 2m and velocity is made 3v

final Kinetic energy (KE')=1/2(2m)(3v)^2

=(1/2mv^2)×18

=KE ×18

therefore KE is increased 18 times

Answer 2

Taking

$ke = \frac{1}{2} m {v}^{2}$

and marking it as equation 1

we take ATQ

$ke = \frac{1}{2} 2m \times ({3v})^{2}$

we get

$\frac{1}{2} 2m \times {9v}^{2}$

=

$\frac{18m {v}^{2} }{2} = 9m {v}^{2}$

therefore kinetic energy becomes 9 times