Question

The kinetic energy of an object of mass, m moving with a velocity of 5 m s
−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

Explanation:

the K.E(KINETIC ENERGY) of the object is 25 J.

So,

25=1/2*m*(5)2

Therefore,m=2 kg.

Now, when the velocity is doubled, velocity becomes=2*5=10 m/s(ms-1)

So,

K.E.=1/2*2*(10)2 =100 J.

Now,when the velocity is increased three times,the velocity will be=3*5=15m/s

Therefore,K.E.=1/2*2*(15)2 = 225 J.

Hope that this may help you sisoo

Answer 2

Given:

$\sf{K_E}{=25J}$

$\sf{Initial \:Velocity =5m}{s}^{-1}$

Solution:

$\sf{K_E=}{\dfrac{1}{2}}{mv}^{2}$

$\sf{25=}{\dfrac{1}{2}}\times{m}\times{25}$

$\sf{m=}{\dfrac{25\times 2}{25}}$

$\sf{m=2kg}$

If it's velocity is doubled:

$\sf{K_E=}{\dfrac{1}{2}}{mv}^{2}$

$\sf{K_E=}{\dfrac{1}{2}}\times{2}\times{100}$

$\sf{K_E=100J}$

If it's velocity is increased three times:

$\sf{K_E=}{\dfrac{1}{2}}{mv}^{2}$

$\sf{K_E=}{\dfrac{1}{2}}\times{2}\times{225}$

$\sf{K_E=225J}$