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IP An archer shoots a 0.024-kg arrow at a target with a speed of \(54 \mathrm{m} / \mathrm{s}\). When it hits the target, it penetrates to a depth of \(0.083 \mathrm{m} .\) (a) What was the average force exerted by the target on the arrow? (b) If the mass of the arrow is doubled, and the force exerted by the target on the arrow remains the same, by what multiplicative factor does the penetration depth change? Explain.

Short Answer

Expert verified
(a) Average force is approximately 422 N. (b) Penetration depth doubles.

Step by step solution

01

Identify the Given Values and Required Quantities

We are given the mass of the arrow, \(m = 0.024\, \text{kg}\), the initial speed \(v_i = 54\, \text{m/s}\), and the penetration depth \(d = 0.083\, \text{m}\). We need to find the average force \(F\) exerted by the target on the arrow.
02

Apply Work-Energy Principle

Using the work-energy principle, the work done on the arrow is equal to the change in kinetic energy. The work done (\(W\)) by the force exerted can be expressed as \(W = F \cdot d\). The initial kinetic energy (\(KE_i\)) is given by \(\frac{1}{2}mv_i^2\), and since the final kinetic energy is zero (arrow stops), the change in kinetic energy is \(-KE_i = -\frac{1}{2}mv_i^2\).
03

Calculate the Average Force Exerted by the Target

Setting the work done equal to the change in kinetic energy, the equation becomes: \[-\frac{1}{2}mv_i^2 = F \cdot d\]Solving for \(F\), we have: \[F = -\frac{\frac{1}{2}mv_i^2}{d}\]Substitute the given values: \[F = -\frac{\frac{1}{2} \cdot 0.024 \cdot 54^2}{0.083}\] Calculate to find \(F\).
04

Analyze Penetration Depth for Doubled Mass

When the mass of the arrow is doubled and the force remains the same, let the new mass be \(m' = 2m\). The new initial kinetic energy \(KE'\) then becomes \(\frac{1}{2}(2m)v_i^2\), which equals \(2 \cdot \frac{1}{2}mv_i^2 = mv_i^2\). The work done is still \(F \cdot d'\). Since \(mv_i^2 = F \cdot d'\), the equation for the new depth is \[d' = \frac{mv_i^2}{F} = 2d\].
05

Determine the Multiplicative Factor

The penetration depth changes by a factor of 2, as doubling the mass doubles the new kinetic energy and thus requires twice the work to stop, leading to twice the penetration depth if the force remains constant.

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is a key concept in physics that describes the energy an object possesses due to its motion. In the context of the archer's arrow, this is calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the arrow. This formula reveals that kinetic energy is directly proportional to both the mass of the object and the square of its velocity.
  • In scenarios like this one, a fast-moving arrow possesses significant kinetic energy, which is transferred upon impact with the target.
  • When the arrow hits, the kinetic energy is used to move the target material to create a path for penetration, or to convert this kinetic energy to another form (like heat or sound).
If we want to stop the arrow completely, all of its initial kinetic energy must be dissipated, aligning with the principles of energy conservation.
Role of Average Force in Stopping the Arrow
The average force exerted by the target on the arrow plays a crucial role in the context of the work-energy principle. This principle states that the work done by an external force on an object translates into changes in the kinetic energy of that object.
In this exercise, the work done by the target's force is used to stop the arrow, meaning that this work equals the loss in kinetic energy as the arrow penetrates and halts. The relationship between force, work, and penetration depth is captured by the equation \( W = F \cdot d \), where \( W \) is the work done and \( d \) is the penetration depth.
  • The force effectively converts or dissipates the arrow's kinetic energy into work, which then stops the arrow.
  • Higher average force would reduce penetration depth, while lower force would increase the depth at the same initial kinetic energy level.
Understanding this concept is necessary to solve for the average force in real-life applications, as it helps to predict material resistance and stopping behaviors.
Exploring Penetration Depth and Its Influences
Penetration depth is the distance an object moves into a material and is directly affected by kinetic energy and the average force exerted. In our problem, the arrow's penetration depth was a result of the work done by the target to stop the arrow. This depth depends on several factors.
  • When the arrow hits the target, all its kinetic energy contributes to penetrating the target to a specific depth.
  • Changes in either mass or velocity alter initial kinetic energy and thus affect penetration depth if the force remains constant.
By understanding the mathematics behind penetration depth, predictions and calculations about how deep objects penetrate can be made more accurately. This is particularly useful in fields requiring material testing or projectile analysis.
Interlinking Mass and Force: Effects on Penetration
The relationship between mass, force, and penetration is an interlocking trio in understanding how projectiles work. If the mass of the arrow is increased while the force exerted by the target remains unchanged, the penetration depth is influenced. For example, doubling the mass of the arrow (as seen in our exercise) changes the depth of penetration without altering the force applied at impact.
  • Doubling the arrow's mass leads to a doubling in kinetic energy (assuming consistent velocity), thus affecting the work needed to be done by the target to stop it.
  • The unchanged force now requires twice the amount of work to halt the now-heavier arrow, consequently doubling the penetration depth. This follows directly from the formula \( d' = \frac{mv_i^2}{F} \), where a doubled \( m \) results in doubled \( d \), if \( F \) remains constant.
This relationship highlights key considerations in both impact physics and material engineering, where both mass and resistance are significant factors in resulting penetration outcomes.

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