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Chapter 3: Problem 56

A \(12.0\) -g bullet is accelerated from rest to a speed of \(700 \mathrm{~m} / \mathrm{s}\) as it travels \(20.0 \mathrm{~cm}\) in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force? [Hint: Be careful with units.

Short Answer

Expert verified
The accelerating force was 14,700 N.

Step by step solution

01

Convert Units

Convert the bullet mass from grams to kilograms and the distance from centimeters to meters. - Mass: \(12.0\) g = \(0.012\) kg - Distance: \(20.0\) cm = \(0.20\) m
02

Use Kinematic Equation to Find Acceleration

Use the kinematic equation \(v^2 = u^2 + 2a s\) to calculate the acceleration. Here:- Final velocity \(v = 700\) m/s- Initial velocity \(u = 0\) m/s (since it starts from rest)- Distance \(s = 0.20\) m- Substitute and solve for acceleration \(a\):\[700^2 = 0 + 2a(0.20)\]\[490000 = 0.4a\]\[a = \frac{490000}{0.4} = 1225000 \text{ m/s}^2\]
03

Calculate the Force Using Newton's Second Law

Using Newton's Second Law, \(F = ma\), find the force:- Mass \(m = 0.012\) kg- Acceleration \(a = 1225000\) m/s\(^2\)- Substitute and solve for the force \(F\):\[F = 0.012 \times 1225000 = 14700 \text{ N}\]

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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 is a fundamental principle in physics that helps us understand how forces affect the motion of objects. It is mathematically expressed as \( F = ma \), where \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced. This equation tells us that the force applied to an object is directly proportional to the acceleration it gains.
In our exercise, the bullet is subjected to an accelerating force as it travels through the gun barrel. By applying Newton鈥檚 Second Law, we determine the size of this force, given the mass and the calculated acceleration. Once we know the bullet's mass and its acceleration, we can easily compute the force needed to achieve its high speed. This demonstrates how interconnected mass, acceleration, and force are in motion scenarios.
Constant Acceleration
Constant acceleration refers to a situation where the acceleration of an object does not change as it moves. This means that every second, the object鈥檚 velocity changes by the same amount. Constant acceleration is a key assumption when using certain kinematic equations.
For our bullet, we assume that the acceleration is constant as it travels down the barrel, which allows us to apply the kinematic equation:
  • \( v^2 = u^2 + 2as \)
Using this formula, we can find the bullet鈥檚 acceleration based on its initial and final velocities and the distance it travels. This equation simplifies calculations and helps predict the behavior of objects under uniform acceleration, making it a powerful tool in physics.
Unit Conversion
Unit conversion is an essential skill in solving physics problems, as it ensures that all quantities are expressed in the proper units for calculation. In physics, the standard units are part of the International System of Units (SI), which includes meters for length, kilograms for mass, and seconds for time.
In our exercise, converting the bullet's mass and the distance it travels into SI units is crucial. We start by changing the mass from grams to kilograms, since the SI unit for mass is kilograms. Similarly, the distance is converted from centimeters to meters. This step prevents errors and ensures the calculations are consistent.
Proper unit conversion is not just a formality but a necessary process that underpins accurate and reliable results in physics computations.

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Most popular questions from this chapter

An unknown force acting on a \(50.0\) -g body floating in space produces a constant acceleration of \(20.0 \mathrm{~cm} / \mathrm{s}^{2}\). If the same force is now made to act on a different body, also in space, producing a constant acceleration of \(40.0 \mathrm{~cm} / \mathrm{s}^{2}\), what is the mass of that body?

The coefficient of static friction between a box and the flat bed of a truck is 0.60. What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{f}=\mu \mathrm{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck; otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{f}=m a_{x}\). However, if the box is on the verge of slipping, \(F_{f}\) \(=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.

Once ignited, a small rocket motor on a spacecraft exerts a constant force of \(10 \mathrm{~N}\) for \(7.80 \mathrm{~s}\). During the burn, the rocket causes the 100 -kg craft to accelerate uniformly. Determine that acceleration.

A \(5.0\) -kg object is to be given an upward acceleration of \(0.30 \mathrm{~m} / \mathrm{s}^{2}\) by a rope pulling straight upward on it. What must be the tension in the rope? The free-body diagram for the object is shown in Fig. \(3-8(b)\). The tension in the rope is \(F_{T}\), and the weight of the object is \(F_{W}=m g=\) \((5.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=49.1 \mathrm{~N}\). Using \(\Sigma F_{y}=m a_{y}\) with up taken as positive, from which \(F_{T}=50.6 \mathrm{~N}=51 \mathrm{~N}\). As a check, we notice that \(F_{T}\) is larger than \(F_{W}\), as it must be if the object is to accelerate upward.

An elevator starts from rest with a constant upward acceleration. It moves \(2.0 \mathrm{~m}\) in the first \(0.60 \mathrm{~s}\). A passenger in the elevator is holding a \(3.0\) -kg package by a vertical string. What is the tension in the string during the accelerating process?
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