A 0.50g bullet is fired at 900m/s in a straight line 1.5m above ground. What is the total mechanical energy of the bullet?
To calculate the total mechanical energy of the bullet, we need to consider both its kinetic energy and its gravitational potential energy.
1. Kinetic energy (KE):
The kinetic energy of an object is given by the equation KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.
Given:
- Mass of the bullet (m) = 0.50 g = 0.50/1000 = 0.00050 kg
- Velocity of the bullet (v) = 900 m/s
Using the formula, the kinetic energy of the bullet is:
KE = 1/2 * 0.00050 kg * (900 m/s)^2
KE = 1/2 * 0.00050 kg * 810,000 m^2/s^2
KE = 205 kg·m^2/s^2 (or J)
2. Gravitational potential energy (PE):
The gravitational potential energy is given by the equation PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height above the reference point (in this case, the ground).
Given:
- Mass of the bullet (m) = 0.00050 kg
- Height above the ground (h) = 1.5 m
- Acceleration due to gravity (g) = 9.8 m/s^2
Using the formula, the gravitational potential energy of the bullet is:
PE = 0.00050 kg * 9.8 m/s^2 * 1.5 m
PE = 0.00735 kg·m^2/s^2 (or J)
3. Total Mechanical Energy (TME):
The total mechanical energy of the bullet is the sum of its kinetic energy (KE) and gravitational potential energy (PE).
TME = KE + PE
TME = 205 kg·m^2/s^2 + 0.00735 kg·m^2/s^2
TME = 205.00735 kg·m^2/s^2 (or J)
Therefore, the total mechanical energy of the bullet is approximately 205.00735 J.