# A gun is fired, and a 50gram bullet is accelerated to a muzzle speed of 100m/s. If the length of the gun barrel is 0.90m, what is the magnitude of the accelerating force? Assume the acceleration is constant.

## acceleration = change in velocity/change in time

magnitude of change in velocity = 100 m/s
(all in one direction)

time = distance/average speed = 0.9/50
= 1.8/100 = .018
so
a = 100/.018
then
F = m a = .050*100/.018 = 278 m/s^2
which is about 28 times gravity

## v² = 2ax

100² = 2*0.9a
a =100² /1.8 = 5555.56
Force = ma = 0.05 x 5555.56 = 278 N

## To find the magnitude of the accelerating force in this scenario, we can use the equation for linear motion:

F = m * a

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

In this case, the mass of the bullet is given as 50 grams, which can be converted to kilograms:

m = 50 grams = 0.05 kg

The acceleration is not directly given, but we can calculate it using the equation of linear motion:

v^2 = u^2 + 2aS

Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
S is the displacement.

In this case, the initial velocity is 0 m/s (as the bullet starts from rest), the final velocity is 100 m/s, and the displacement is given as the length of the gun barrel, which is 0.90 m. Rearranging the equation, we get:

a = (v^2 - u^2) / (2S)

Substituting the values, we get:

a = (100^2 - 0^2) / (2 * 0.90)

Now we can substitute the calculated mass and acceleration in the previous equation to find the force:

F = m * a

F = 0.05 kg * [(100^2 - 0^2) / (2 * 0.90)]

Simplifying the equation, we get:

F = 0.05 kg * [(10000) / (1.8)]

Finally, calculating the value, we find that the magnitude of the accelerating force is approximately 2777.78 Newtons (N).

F ≈ 2777.78 N