Answer to Question #206083 in Physics for Mya

Question

A rifle can shoot a 3.8 x 10^-2kg bullet at the speed of 790 m/s

Calculate the kinetic energy of the bullet as it leaves the rifle?

When that is solved the next question is if the bullet is 2 meters above the ground at that moment, what is the potential energy?

Answer 1

Kinetic energy of the bullet when it leaves the rifle is $E_K = \frac{m v^2}{2} = \frac{3.8 \cdot 10^{-2} kg \cdot (790 m/s)^2}{2} \approx 11857.9 J$.

Potential energy of the bullet is $E_P = m g h = 3.8 \cdot 10^{-2} kg \cdot 9.81 \frac{m}{s^2} \cdot 2 m \approx 0.75 J$.