# What is the kinetic energy in joules of a 0.02 kg bullet traveling 300 m/s?

Mar 18, 2018

$900 J$

#### Explanation:

Kinetic energy of bullet = $K . E = \frac{1}{2} m {v}^{2}$ where, $m$ is the mass of the bullet and $v$ is its velocity .

Given, $m = 0.02 K g , v = 300 m {s}^{-} 1$

So,$K . E = \frac{1}{2} \cdot \left(0.02\right) {\left(300\right)}^{2} = 900 J$

Mar 18, 2018

$900 \setminus \text{J}$

#### Explanation:

Well, kinetic energy is expressed through the equation:

$\text{KE} = \frac{1}{2} m {v}^{2}$

• $m$ is the mass of the object in kilograms

• $v$ is the velocity of the object in meters per second

So, we plug in the values, which are

$m = 0.02 \setminus \text{kg}$

$v = 300 \setminus \text{m/s}$

And we get,

"KE"=1/2*0.02 \ "kg"*(300 \ "m/s")^2

$= \frac{1}{2} \cdot 0.02 \setminus {\text{kg"*90000 \ "m"^2"/s}}^{2}$

$= 900 \setminus {\text{kg m"^2"/s}}^{2}$

$= 900 \setminus \text{J}$