A 7.74- g bullet is moving at 578.00 m/s as it leaves the 0.64- m-long barrel of a rifle. What is the average force on the bullet as it moves down the barrel? Assume that the acceleration is constant.
s = v^2/2a
solve that for a, and the F=ma.
To find the average force on the bullet as it moves down the barrel, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's determine the acceleration of the bullet. We can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (578.00 m/s)
u = initial velocity (0 m/s)
a = acceleration (unknown)
s = displacement (0.64 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the values into the formula:
a = (578.00^2 - 0^2) / (2 * 0.64)
Next, we need to calculate the mass of the bullet in kilograms. The given mass is 7.74 g, so we convert it to kilograms:
m = 7.74 g / 1000
Putting the values back into the original formula, we can now calculate the average force:
Average force = mass * acceleration
Substituting the values:
Average force = (7.74/1000) * [(578.00^2 - 0^2) / (2 * 0.64)]
Evaluating the expression will give you the average force on the bullet as it moves down the barrel.