A 5.8 g bullet leaves the muzzle of a rifle with a speed of 316 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.83 m long barrel of the rifle?
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To determine the force exerted on the bullet while it is traveling down the barrel of the rifle, we can make use of Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a).
In this case, the mass of the bullet is given as 5.8 grams, which can be converted to kilograms by dividing by 1000: 5.8 g ÷ 1000 = 0.0058 kg.
The acceleration of the bullet can be determined using the formula for average acceleration, which is the change in velocity (Δv) divided by the change in time (Δt). However, since the problem does not provide information about the time it takes for the bullet to travel down the barrel, we can use an alternative approach to determine the acceleration.
The final velocity of the bullet (v) is given as 316 m/s, and the initial velocity (u) is assumed to be zero since the bullet starts from rest. Thus, the change in velocity is Δv = v - u = 316 m/s - 0 m/s = 316 m/s.
Now, we can use another kinematic equation that relates velocity, acceleration, and displacement: v^2 = u^2 + 2as. In this case, the initial velocity (u) and displacement (s) are both zero since the bullet starts from rest at the muzzle of the rifle. Therefore, the equation simplifies to v^2 = 2as.
Rearranging the equation to solve for acceleration (a), we have a = v^2 / (2s).
Plugging in the values given, we have a = (316 m/s)^2 / (2 * 0.83 m).
Calculating this, we find that the acceleration of the bullet is approximately 6075.301 m/s^2.
Finally, we can calculate the force exerted on the bullet using Newton's second law: F = m * a.
Plugging in the values, we have F = 0.0058 kg * 6075.301 m/s^2.
Evaluating this expression, we find that the force exerted on the bullet while it is traveling down the barrel is approximately 35.25 N.