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Chapter 5: Problem 30

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13 \mathrm{~g}\), its speed before entering the branch was \(220 \mathrm{~m} / \mathrm{s},\) and its penetration depth was \(15 \mathrm{~mm} .\) If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

Short Answer

Expert verified
The force of the branch on the toothpick was 209.734 N.

Step by step solution

01

Convert Mass to Kilograms

The mass of the toothpick is given in grams. First, we need to convert it to kilograms for our calculations.Given: \[m = 0.13 \text{ g} = 0.13 \times 10^{-3} \text{ kg} = 0.00013 \text{ kg}\]
02

Identify Initial and Final Velocities

The initial speed \(v_i\) of the toothpick is given as \(220 \text{ m/s}\). The final speed \(v_f\) is \(0 \text{ m/s}\) because the toothpick stops upon penetration.
03

Determine the Penetration Depth

The penetration depth \(d\) is given as \(15 \text{ mm}\). Convert this to meters:\[d = 15 \text{ mm} = 15 \times 10^{-3} \text{ m} = 0.015 \text{ m}\]
04

Use Kinematic Equation to Find Deceleration

We will use the kinematic equation:\[v_f^2 = v_i^2 + 2a d\]Plug in the known values:\[0 = (220)^2 + 2a(0.015)\]Solve for \(a\):\[a = -\frac{(220)^2}{2 \times 0.015} = -1613333.33 \text{ m/s}^2\]
05

Calculate Force Using Newton's Second Law

Newton's second law relates force, mass, and acceleration as:\[F = ma\]Substitute the known mass and calculated acceleration:\[F = 0.00013 \times (-1613333.33) = -209.734 \text{ N}\]Magnitude of the force is \[209.734 \text{ N}\]
06

Determine Magnitude of Force

Since we're interested in the magnitude and not the direction, take the absolute value of the force computed:\[\text{Magnitude of Force} = |F| = 209.734 \text{ N}\]

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Key Concepts

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

Kinematic Equations
Kinematic equations are mathematical equations that describe the motion of objects. They are essential when analyzing scenarios where an object moves at a uniform acceleration. In our exercise, a toothpick is shot into an oak branch and comes to a stop. This scenario can be analyzed using one of the fundamental kinematic equations: \[ v_f^2 = v_i^2 + 2a d \].
  • **Initial Velocity (\(v_i\)):** The speed of the toothpick before it hits the branch, which is 220 m/s.
  • **Final Velocity (\(v_f\)):** Since the toothpick stops, its final velocity is 0 m/s.
  • **Penetration Depth (\(d\)):** The distance the toothpick travels while stopping, which is converted from millimeters to meters as 0.015 m.
  • **Acceleration (\(a\)):** Using the kinematic equation, we solve for \(a\), which represents the deceleration of the toothpick.
By substituting these values, we can determine how quickly the toothpick decelerated.
Deceleration
Deceleration is essentially negative acceleration, indicating a decrease in an object's speed. When the toothpick enters the oak branch, it slows down dramatically until it stops.In the context of our exercise, we used the appropriate kinematic equation that contains acceleration to find the deceleration. The equation, \[ v_f^2 = v_i^2 + 2a d \]was rearranged to solve for \(a\):\[ a = -\frac{(220)^2}{2 \times 0.015} \].Here, the initial velocity of 220 m/s declines to 0 over a 0.015 m distance, giving us a very large deceleration value of approximately \[-1613333.33 \, \text{m/s}^2\].This large number signifies the rapid rate of deceleration when the toothpick enters the branch, illustrating the substantial force exerted by the branch during the process.
Force Calculation
To calculate the force exerted on the toothpick, we turn to Newton's Second Law. This fundamental law of motion states that the force acting on an object is the product of its mass and acceleration, given by the formula:\[ F = ma \].In the exercise:
  • The **mass** of the toothpick is \[0.00013 \, \text{kg}\], having been converted from grams.
  • The **acceleration** calculated earlier is actually a deceleration at \[-1613333.33 \, \text{m/s}^2\].
When these values are substituted into the formula:\[ F = 0.00013 \times (-1613333.33) \].This yields a force of \[-209.734 \, \text{N}\].However, because we are interested in the magnitude of this force, we consider the absolute value, resulting in \[209.734 \, \text{N}\].This force magnitude reflects how intensely the branch resists the fast-moving toothpick, achieving a complete stop within a short distance.

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Most popular questions from this chapter

A certain particle has a weight of \(22 \mathrm{~N}\) at a point where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2} .\) What are its (a) weight and (b) mass at a point where \(g=4.9 \mathrm{~m} / \mathrm{s}^{2} ?\) What are its (c) weight and (d) mass if it is moved to a point in space where \(g=0 ?\)

An electron with a speed of \(1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) moves horizontally into a region where a constant vertical force of \(4.5 \times 10^{-16} \mathrm{~N}\) acts on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg} .\) Determine the vertical distance the electron is deflected during the time it has moved \(30 \mathrm{~mm}\) horizontally.

The tension at which a fishing line snaps is commonly called the line's "strength." What minimum strength is needed for a line that is to stop a salmon of weight \(85 \mathrm{~N}\) in \(11 \mathrm{~cm}\) if the fish is initially drifting at \(2.8 \mathrm{~m} / \mathrm{s}\) ? Assume a constant deceleration.

In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about \(42^{\circ}\) ) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a \(7.260 \mathrm{~kg}\) shot is accelerated along a straight path of length \(1.650 \mathrm{~m}\) by a constant applied force of magnitude \(380.0 \mathrm{~N}\), starting with an initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) \(30.00^{\circ}\) and (b) \(42.00^{\circ}\) ? (Hint: Treat the motion as though it were along a ramp at the given angle. \()\) (c) By what percent is the launch speed decreased if the athlete increases the angle from \(30.00^{\circ}\) to \(42.00^{\circ} ?\)

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m} .\) He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)
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