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A \(5.0-g\) bullet leaves the muzzle of a rifle with a speed of \(320 \mathrm{~m} / \mathrm{s}\). What force (assumed constant) is exerted on the bullet while it is traveling down the \(0.82\)-m-long barrel of the rifle?

Short Answer

Expert verified
The force exerted on the bullet while it is traveling down the barrel of the rifle is approximately \(312.20N\).

Step by step solution

01

Determine the bullet's acceleration

Using the equation \(v^2 = u^2 + 2as\) which describes the relationship between displacement, initial and final velocities and acceleration. Here, \(u = 0m/s\) (bullet starts from rest inside the barrel), \(v = 320m/s\) (final speed of the bullet), and \(s = 0.82m\) (length of the barrel). Rearrange the equation to solve for a: \(a = (v^2 - u^2) / 2s\). Substituting the given values, \(a = (320^2 - 0^2) / (2*0.82)\). Calculate the value of this gives \(a = 62439.024m/s^2\).
02

Calculate the force exerted on the bullet using Newton's second law

According to Newton's second law, the force exerted on an object is calculated by the formula: \(F=m*a\), Where \(m\) is the mass of the object and \(a\) is the acceleration, Here, \(m = 5g = 0.005kg\) (converting grams to kilograms), and \(a\) from step 1 is \(62439.024m/s^2\). So, \(F = 0.005 * 62439.024\) which gives \(F = 312.19512N\). Thus, the force exerted on the bullet while it is traveling down the barrel of the rifle is approximately \(312.20N\) (rounded to two decimal places).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law of Motion is a fundamental principle in physics that explains how the motion of an object is affected by the forces acting on it. According to this law, the force exerted on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it is expressed as:

\( F = m \cdot a \)

Where:
  • \(F\) is the force applied (in Newtons)
  • \(m\) is the mass (in kilograms)
  • \(a\) is the acceleration (in meters per second squared)
This relationship tells us that the greater the mass of an object, the more force is required to change its motion. Similarly, an increase in acceleration leads to a higher force for a given mass. In the context of our exercise, when the bullet is fired, we calculate the force exerted on it by first determining its acceleration as it travels through the rifle barrel. By applying Newton's Second Law, we find how much force the rifle applies to the bullet to achieve its final speed.
Projectile Motion
Projectile motion refers to the phenomenon where an object moves in a curved trajectory under the influence of gravity alone. This kind of motion is characterized by two components, horizontal and vertical, that happen simultaneously. Although the bullet in our example primarily travels in a straight line within the rifle barrel, once it exits, it may enter projectile motion due to gravity's influence.

Key aspects of projectile motion include:
  • Independent horizontal and vertical motions
  • A constant horizontal velocity if we assume no air resistance
  • An acceleration in the vertical motion due to gravity
Understanding projectile motion is crucial for analyzing how objects behave after being set in motion with an initial velocity. To predict a projectile's path, such as a bullet's trajectory after exiting the barrel, one must consider both components together. Although the exercise focuses on the force inside the barrel, extending this knowledge to how the bullet behaves outside adds depth to your understanding of motion dynamics.
Acceleration Calculation
Acceleration is the rate at which an object's velocity changes with time. When calculating acceleration, it is important to know several key variables, like the initial and final velocities, as well as the distance over which the change occurs. In our exercise, to find the bullet's acceleration, we use the kinematic equation:

\[ v^2 = u^2 + 2as \]

Here:
  • \(v\) is the final velocity
  • \(u\) is the initial velocity
  • \(a\) is the acceleration
  • \(s\) is the displacement
Rearranging the equation helps solve for acceleration \(a\) as follows:

\[ a = \frac{v^2 - u^2}{2s} \]

Substitute the values with the bullet starting from rest (\(u = 0\)) and traveling down a barrel of known length (\(s = 0.82\, m\)). The final velocity is \(320 \, m/s\). Solving for \(a\) gives a remarkable acceleration value, which illustrates how swiftly forces act on the bullet. Mastering such calculations allow us to evaluate not just bullets, but any object experiencing constant acceleration, enhancing our comprehension of motion principles in physics.

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Most popular questions from this chapter

(a) An elevator of mass \(m\) moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, \(T\) or \(w ?\) (b) When the elevator is moving at a constant velocity upward, which is greater, \(T\) or \(w\) ? (c) When the elevator is moving upward, but the acceleration is downward, which is greater, \(T\) or \(w ?\) (d) Let the elevator have a mass of \(1500 \mathrm{~kg}\) and an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of \(10 \mathrm{~m} / \mathrm{s}\). Find T. Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with your answer to part (c)?

A block of mass \(m=\) \(5.8 \mathrm{~kg}\) is pulled up a \(\theta=\) \(25^{\circ}\) incline as in Figure P4.30 with a force of magnitude \(F=32 \mathrm{~N}\). (a) Find the acceleration of the block if the incline is frictionless. (b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is \(0.10\).

The distance between two telephone poles is \(50.0 \mathrm{~m}\). When a \(1.00-\mathrm{kg}\) bird lands on the telephone wire midway between the poles, the wire sags \(0.200 \mathrm{~m}\). Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.

A fisherman poles a boat as he searches for his next catch. He pushes parallel to the length of the light pole, exerting a force of \(240 \mathrm{~N}\) on the bottom of a shallow lake. The pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of \(35.0^{\circ}\) with the vertical and the water exerts a horizontal drag force of \(47.5 \mathrm{~N}\) on the boat, opposite to its forward velocity of magnitude \(0.857 \mathrm{~m} / \mathrm{s}\). The mass of the boat including its cargo and the worker is \(370 \mathrm{~kg}\). (a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force. (b) Assume the forces are constant over a short interval of time. Find the velocity of the boat \(0.450 \mathrm{~s}\) after the moment described. (c) If the angle of the pole with respect to the vertical increased but the exerted force against the bottom remained the same, what would happen to buoyant force and the acceleration of the boat?

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force \(F\) exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of \(1.50 \mathrm{~m}\) and a mass of \(61.0 \mathrm{~kg}\). If this fish is moving upward at \(3.00 \mathrm{~m} / \mathrm{s}\) as its head first breaks the surface and has an upward speed of \(6.00 \mathrm{~m} / \mathrm{s}\) after two-thirds of its length has left the surface, assume constant acceleration and determine (a) the salmon's acceleration and (b) the magnitude of the force \(F\) during this interval.

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