**Kinetic Energy**

Hey students we already learned that, ‘the energy possessed by a body due to its motion is known as kinetic energy’. E.g. Athletes running on a track possess kinetic energy, bullet fired from gun is also an example of kinetic energy. So let’s derive the formula for kinetic energy of a body/object.

Imagine a boy of mass ‘m’ starts from rest and runs on running track at speed ‘v’. Let ‘t’ be time required by him to cover the distance ‘s’.

Let ‘t’ be time required by him to cover the distance ‘s’.

a= acceleration of boy while running

then the work done by him will be,

W = F.s ……………..(1)

But by Newton’s 2nd law of motion, we have

F = ma ……….(2)

Then equation (1) becomes

W = m × a × s ………………….(3)

We have by third equation of motion,

v^{2} = u^{2} + 2as

∴2as = v^{2} – u^{2}

∴ a = v^{2 }– u^{2} / 2s

Then equation (3) becomes

We know that this work done is always equal to the energy gained/spent by body i.e. this work is nothing but the kinetic energy,

∴ W = E = 1/2 mv^{2}

∴ Kinetic energy of body = 1/2 mv^{2}

SI unit of kinetic energy is joule (J).

CGS unit of kinetic energy is erg. Remember that 1 joule = 10^{7} erg

Following are some numeric which will give us more details about kinetic energy.

**Ex:1) A bullet of mass 100 g fired with velocity of 250 cm/s. Find the kinetic energy is joule.**

Ans: Here m=100 g

v = 250 cm/s

The formula for kinetic energy is

E = ½ mv^{2}

∴E = ½ × 100 × 250^{2}

∴ E = 50 × 250 × 250

∴ E = 31,25,000 erg

∴ E = 0.3125 joule

**Ex:2) Energy of 25 J is required to throw a cricket ball of mass 0.5 kg from boundary to wicket keeper, find the velocity required to reach the ball to wicket keeper?**

Ans: Here m = 0.5 g

E = 25 J

v = ? m/s

The formula for kinetic energy is

E = ½ mv^{2}

∴ 25 = ½ × 0.5 × v^{2}

∴ 50 = 0.5 × v^{2}

∴ v^{2} = 50/0.5

∴ v^{2} = 100

∴ v = 10 m/s