Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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?

seo-qna
Last updated date: 16th Jul 2024
Total views: 347.7k
Views today: 8.47k
Answer
VerifiedVerified
347.7k+ views
Hint:We know that the kinetic energy is expressed as the energy which is due to the motion of any object. It depends on the mass and the velocity of the moving object. here, we will find the correlation between the velocity and the energy of an object and will find the mass of the body.

Complete step by step answer:
As the velocity is given as $5\,m{s^{ - 1}}$and Kinetic energy is given as $25\,J$, we can use the above equation and find out its mass.
$K.E = \dfrac{1}{2}m{v^2}$
Putting in the values of K.E and velocity we will get,
$25 = \dfrac{1}{2}m{(5)^2}$
Rearranging the above relation to calculate the mass, we will get,
$\dfrac{{25 \times 2}}{{{5^2}}} = m$
Solving the above equation will give us mass as,
$m = 2kg$
Now if velocity is doubled, i.e. $v = 10\,m{s^{ - 1}}$, K.E. will be,
$K.E = \dfrac{1}{2}m{v^2}$
Substituting the value of v and m in order to calculate the K.E. will give us,
$K.E = \dfrac{1}{2} \times 2 \times {10^2}$
Solving the above equation will give us,
$\therefore K.E = 100\,J$
Now if velocity is three times, i.e. $v = 15\,m{s^{ - 1}}$, K.E. will be
$K.E = \dfrac{1}{2}m{v^2}$
Substituting the value of v and m in order to calculate the K.E. will give us,
$K.E = \dfrac{1}{2} \times 2 \times {15^2}$
Solving the above equation will give us,
$\therefore K.E = 225\,J$

Hence, its kinetic energy when its velocity is doubled is 100 J and its kinetic energy when its velocity is increased three times is 225 J.

Note:The alternative method to find the solution of the given example can be done as we can also see that the mass ‘m’ of the object was constant so we can also directly calculate the K.E. by the relation,
$K.E \propto {v^2}$
where, $v = 5$
So when velocity is doubled we can write it as,
$K.E \propto {(2v)^2}$
$\Rightarrow K.E \propto 4{v^2}$
$\Rightarrow K.E = 4 \times 25$
$\therefore K.E = 100\,J$
Now, when velocity is three times we can write it as,
$K.E \propto {(3v)^2}$
$\Rightarrow K.E \propto 9{v^2}$
$\Rightarrow K.E = 9 \times 25$
$\therefore K.E = 225\,J$