When an 8-lb rifle is fired, the 0.02-lb bullet receives an acceleration of 10^5 ft/s^2 while it is the barrel? (a) How much force acts on the bullet? (b) Does any force act on the rifle? If so, how much and in what direction? (c) The bullet is accelerated for 0.007 s. How fast does it leave the barrel of the rifle?
(a) f = m a = .02/32 * 1E5 = 63 lb
(b) yes ... Newton's 3rd Law ... equal and opposite reaction
(c) .007 s * 1E5 ft/s² = 700 ft/s
sounds like an air rifle
To answer your questions, we'll need to use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration: F = m * a.
(a) First, let's find the force acting on the bullet. The mass of the bullet is 0.02 lb, which is equivalent to approximately 0.0091 kg (1 lb ≈ 0.4536 kg). The acceleration is given as 10^5 ft/s^2, which is approximately 3.048 × 10^4 m/s^2 (1 ft ≈ 0.3048 m). Let's plug these values into the formula:
F = m * a
F = 0.0091 kg * 3.048 × 10^4 m/s^2
F ≈ 0.2776 N
Therefore, the force acting on the bullet is approximately 0.2776 Newtons.
(b) Yes, a force acts on the rifle as well, due to the principle of action and reaction. This force is equal in magnitude but opposite in direction to the force acting on the bullet. Therefore, the force acting on the rifle is also approximately 0.2776 Newtons, but in the opposite direction.
(c) To determine the velocity of the bullet as it leaves the barrel, we can use the equation of motion: v = u + a * t, where v is the final velocity, u is the initial velocity (assumed to be 0 since the bullet starts from rest), a is the acceleration, and t is the time.
Plugging in the values:
v = 0 + 3.048 × 10^4 m/s^2 * 0.007 s
v ≈ 213.36 m/s
Therefore, the bullet leaves the barrel with an approximate velocity of 213.36 m/s.
To answer this question, we can apply 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.
(a) To find the force acting on the bullet, we need to calculate the product of the mass of the bullet and its acceleration. Given that the mass of the bullet is 0.02 lb and the acceleration is 10^5 ft/s^2, we can convert the mass to slugs (lb/s^2) and calculate:
Force = mass × acceleration
Force = (0.02 lb) × (32.174 ft/s^2/lb) × (10^5 ft/s^2)
Force = 643.48 lb·ft/s^2
Therefore, the force acting on the bullet is 643.48 lb·ft/s^2.
(b) According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Therefore, when the bullet is fired forward, the rifle experiences a recoil force in the opposite direction. The magnitude of this force is equal to the force acting on the bullet. Thus, the force acting on the rifle would also be 643.48 lb·ft/s^2, but in the opposite direction.
(c) To calculate the velocity of the bullet as it leaves the barrel, we can use the formula:
final velocity = initial velocity + (acceleration × time)
The initial velocity of the bullet is assumed to be zero since it starts from rest inside the barrel, and the acceleration is 10^5 ft/s^2. Given that the bullet is accelerated for 0.007 s, we can substitute these values into the equation:
final velocity = 0 + (10^5 ft/s^2 × 0.007 s)
final velocity = 700 ft/s
Therefore, the bullet will leave the barrel of the rifle with a velocity of 700 ft/s.