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A \(30 \mathrm{~g}\) bullet moving a horizontal velocity of \(500 \mathrm{~m} / \mathrm{s}\) comes to a stop \(12 \mathrm{~cm}\) within a solid wall. (a) What is the change in the bullet's mechanical energy? (b) What is the magnitude of the average force from the wall stopping it?

Short Answer

Expert verified
The change in mechanical energy is -3750 J. The average force is 31250 N.

Step by step solution

01

Understand the Problem

The bullet's mass is given as 30 g, which is 0.03 kg in SI units. Its initial velocity is 500 m/s, and it comes to rest within a distance of 12 cm or 0.12 m. We are asked to find the change in mechanical energy and the average force exerted by the wall to stop the bullet.
02

Convert Values and Set Variables

Convert the mass of the bullet from grams to kilograms: \( m = 30 \mathrm{~g} = 0.03 \mathrm{~kg} \). The initial velocity is \( v_i = 500 \mathrm{~m/s} \) and final velocity is \( v_f = 0 \mathrm{~m/s} \). The stopping distance is \( d = 12 \mathrm{~cm} = 0.12 \mathrm{~m} \).
03

Calculate Initial Kinetic Energy

Use the kinetic energy formula: \( KE = \frac{1}{2} mv^2 \). The initial kinetic energy \( KE_i = \frac{1}{2} \times 0.03 \times (500)^2 \). Calculate this value.
04

Compute Initial Kinetic Energy Result

Substitute the values: \( KE_i = \frac{1}{2} \times 0.03 \times 250000 = 3750 \mathrm{~J} \).
05

Determine Final Kinetic Energy

The final kinetic energy \( KE_f \) is zero since the bullet comes to a stop: \( KE_f = \frac{1}{2} \times 0.03 \times (0)^2 = 0 \mathrm{~J} \).
06

Find Change in Mechanical Energy

The change in mechanical energy is the difference between initial and final kinetic energies: \( \Delta E = KE_f - KE_i = 0 - 3750 = -3750 \mathrm{~J} \). The negative sign indicates a loss of energy.
07

Apply Work-Energy Principle to Find Average Force

The work done by the force is equal to the change in kinetic energy, which is \( -3750 \mathrm{~J} \). The work done by a constant force is \( F \times d \). Thus, \( F \times 0.12 = 3750 \).
08

Calculate Average Force Magnitude

Solve for \( F \): \( F = \frac{3750}{0.12} = 31250 \mathrm{~N} \). This is the magnitude of the average force exerted by the wall.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a measure of the energy that an object possesses due to its motion. It is calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. The energy is directly proportional to both the mass and the square of the velocity, meaning that as either the mass or velocity increases, the kinetic energy increases significantly.

In the original exercise, the bullet initially moves with a velocity of 500 m/s and has a mass of 0.03 kg. By applying the kinetic energy formula, we can determine the bullet's initial kinetic energy before it hits the wall. This calculation is crucial to understanding how much energy the bullet has and what happens to that energy when it stops.

To calculate the initial kinetic energy, substitute the mass and initial velocity into the formula: \( KE_i = \frac{1}{2} \times 0.03 \times (500)^2 \). Solving this, the bullet starts with an energy of 3750 J (Joules). Understanding kinetic energy is fundamental when analyzing how the bullet's energy changes after collision or any other action that affects its velocity.
Average Force
The concept of force is central to understanding the bullet's transition from motion to rest. When the bullet hits the wall, it experiences an average force that causes it to decelerate to a stop. This force can be described using the work done on the bullet, which is linked to the energy change and the distance over which it acts.

The average force \( F \) exerted by the wall is calculated by applying the work-energy principle: the work done by the force is equal to the energy change. For a force applied over a known distance \( d \), the relationship can be written as \( F \times d = \Delta KE \), where \( \Delta KE \) is the change in kinetic energy. In this case, the total mechanical energy loss is 3750 J, and the bullet stops over a distance of 0.12 m. Solving \( F = \frac{3750}{0.12} \) gives the average force magnitude of 31250 N.

This force is surprisingly large due to the short stopping distance, highlighting how rapidly the wall needs to exert force to stop the bullet. Understanding the average force gives us insight into how objects like bullets interact with surfaces, influencing safety engineering and material science.
Work-Energy Principle
The work-energy principle is a fundamental concept that connects the work done by forces to changes in kinetic energy. Simply put, it states that the work done on an object by all the forces acting on it is equal to the change in its kinetic energy.

In our exercise, this principle helps explain how the bullet's kinetic energy of 3750 J is dissipated as it embeds into the wall. The work done by the wall against the bullet's motion, effectively stopping it, is equal to the negative of this initial energy. The negative sign indicates an energy loss—a decrease in the system's kinetic energy—which aligns with the bullet coming to rest.

By applying this principle, we also see that the stopping force exerted by the wall is directly calculated using the equation \( F \times d = \Delta KE \). This illustrates how work, calculated through force and distance, directly reflects the energy transfer during the bullet’s deceleration. Understanding the work-energy principle is crucial for solving problems involving any object in motion coming to a halt or changing its speed due to an external force.

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Most popular questions from this chapter

Next, the particle is released from rest at \(x=0 .\) What are (l) its kinetic energy at \(x=5.0 \mathrm{~m}\) and \((\mathrm{m})\) the maximum positive position \(x_{\max }\) it reaches? (n) What does the particle do after it reaches \(x_{\max } ?\) $$ \begin{array}{lc} {\text { Range }} & \text { Force } \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \end{array} $$ For the arrangement of forces in Problem \(81,\) a \(2.00 \mathrm{~kg}\) particle is released at \(x=5.00 \mathrm{~m}\) with an initial velocity of \(3.45 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis. (a) If the particle can reach \(x=0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive \(x\) direction when it is released at \(x=5.00 \mathrm{~m}\) at speed \(3.45 \mathrm{~m} / \mathrm{s} .\) (b) If the particle can reach \(x=13.0 \mathrm{~m},\) what is its speed there, and if it cannot, what is its turning point?

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed- squared. For a certain car with a weight of \(12000 \mathrm{~N},\) the total resistant force \(F\) is given by \(F=300+1.8 v^{2}\) with \(F\) in newtons and \(v\) in meters per second. Calculate the power (in horsepower) required to accelerate the car at \(0.92 \mathrm{~m} / \mathrm{s}^{2}\) when the speed is \(80 \mathrm{~km} / \mathrm{h}\)

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A \(50 \mathrm{~g}\) ball is thrown from a window with an initial velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and (b) its speed when it is \(3.0 \mathrm{~m}\) below the window. Does the answer to (b) depend on either (c) the mass of the ball or (d) the initial angle?

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