what is the minimum amount of energy required to accelerate horizontally a 3.0g bullet from rest to a speed of 40.0 m/s?
what is 1/2 mass*velocity^2
so is it 1/2(3)(40)^2?
no, no, no. mass has to be in Kilograms.
oh right. so 1/2(.003)(40)^3
To calculate the minimum amount of energy required to accelerate a bullet from rest to a specific speed, we need to use the principle of conservation of mechanical energy.
The bullet's initial kinetic energy is zero because it starts from rest. The final kinetic energy is given by the formula:
\( KE = \frac{1}{2}mv^2 \)
Where:
KE = kinetic energy
m = mass of the bullet
v = final velocity of the bullet
Given the mass of the bullet, m = 3.0g = 0.003kg, and the final velocity, v = 40.0 m/s, we can substitute these values into the formula:
\( KE = \frac{1}{2}(0.003kg)(40.0m/s)^2 \)
Simplifying the equation:
\( KE = \frac{1}{2}(0.003kg)(1600.0\,m^2/s^2) \)
\( KE = (0.0024kg)(1600.0\,m^2/s^2) \)
\( KE = 3.84\,J \)
Therefore, the minimum amount of energy required to accelerate the 3.0g bullet from rest to a speed of 40.0 m/s is 3.84 Joules.