Question

The kinetic energy of an object of mass 'm' moving with a velocity of 5 m/s is 25 J. what will be its kinetic energy when its velocity is increased three times.​

Answer 1

Answer:

If the velocity of an object is doubled, then its kinetic energy becomes 4 times the original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.

Answer 2

Given:

Kinetic energy of an object of mass m = 25J

Object velocity , v = 5 m/s

To Find :

The Kinetic energy when object velocity is increased three times .

Solution :

Let the intital kinetic energy of object be $\sf\:K_1=25JK$

=25J and initial velocity$\sf\:v_1=v=5ms^{-1}v$

and $\sf\:k_2k$

be kinetic energy of an object when it's Velocity $\sf\:v_2=3v_1=15ms^{-1}v </p><p>2</p><p> </p><p> =3v </p><p>1</p><p> </p><p> =15ms </p><p>−1$

Initial Kinetic energy

$\rm\:K_1=\dfrac{1}{2}m(v_1)^2K$

1

$\sf\implies\:K_1=\dfrac{1}{2}mv^2..(1)⟹K </p><p>1</p><p> </p><p> = </p><p>2</p><p>1</p><p> </p><p> mv </p><p>2</p><p> ..(1)$

Kinetic energy when Velocity increase three times

$\rm\:K_2=\dfrac{1}{2}m(v_2)^2K$

$\sf\implies\:K_2=\dfrac{1}{2}m(3v)^2..(2)⟹K$

Now ,

Divide Equation (1) & (2)

$\sf\dfrac{K_1}{K_2}=\dfrac{\frac{1}{2}mv^2}{\frac{1}{2}m(3v)^2}$

$\sf\implies\dfrac{K_1}{K_2}=\dfrac{1}{9}⟹$

$\sf\implies\dfrac{25}{K_2}=\dfrac{1}{9}⟹$

$\sf\implies\:K_2=25\times9⟹K$

$\sf\implies\:K_2=225J⟹K$

=225J