Question

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

A. 9.00×10³ N
B. 7.50×10³ N
C. 8.00×10³ N
D. 6.50×10³ N

Answer 1

The average force exerted on the bullet is 9.00 × 10³ N.

Step-by-step explanation:

The average force exerted on a bullet can be determined using Newton's second law of motion:

F = ma

Where F is the force, m is the mass, and a is the acceleration.

We can rearrange the equation to solve for the force:

F = ma = m * (Δv / Δt)

Given that the mass of the bullet is 0.0300 kg, the change in velocity (Δv) is 600.0 m/s, and the time (Δt) is 2.00 ms (or 0.002 s), we can substitute the values into the equation to calculate the force:

F = (0.0300 kg) * (600.0 m/s / 0.002 s) = 9.00 × 10³ N

Therefore, the average force exerted on the bullet is 9.00 × 10³ N.