/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 The high-speed winds around a to... [FREE SOLUTION] | 91影视

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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 magnitude of the force is 209.73 N.

Step by step solution

01

Convert Units

First, convert the mass of the toothpick from grams to kilograms. Since there are 1000 grams in a kilogram, \(0.13\, \text{g} = 0.00013\, \text{kg}\). Next, convert the penetration depth from millimeters to meters. Since there are 1000 millimeters in a meter, \(15 \text{ mm} = 0.015 \text{ m}\).
02

Calculate Initial Kinetic Energy

Calculate the initial kinetic energy \( K_i \) of the toothpick using the formula \( K_i = \frac{1}{2} m v^2 \), where \( m = 0.00013 \) kg and \( v = 220 \text{ m/s} \). Substitute the values to get: \( K_i = \frac{1}{2} \times 0.00013 \times 220^2 = 3.146 \text{ J}\).
03

Calculate Work Done by the Force

Since the force from the branch is the only force acting on the toothpick (assuming other forces like gravity are negligible during penetration), the work-energy principle gives \( W = \Delta KE = - K_i \), because the final kinetic energy \( K_f = 0 \). Thus, \( W = -3.146 \text{ J}\).
04

Calculate the Force

Use the formula for work, \( W = F \cdot d \), where \( F \) is the force, and \( d \) is the penetration depth \(0.015 \text{ m}\). Substitute \( W = -3.146 \text{ J}\) and \( d = 0.015 \text{ m}\) into this equation to solve for \( F \): \( F = \frac{-3.146}{0.015} = -209.73 \text{ N}\). Since we need the magnitude of the force, \( F = 209.73 \text{ N}\).

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Key Concepts

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

Kinetic Energy
When we talk about kinetic energy, we are talking about the energy that an object possesses due to its motion. Kinetic energy is an essential concept in physics and it's calculated using the formula:\[ K = \frac{1}{2} m v^2 \]Here, *m* stands for mass and *v* is the velocity of the object. When a toothpick moves at high speed into an oak branch, it possesses kinetic energy. The faster an object moves, or the greater its mass, the more kinetic energy it will have. This energy is crucial because it's what gets transferred when moving objects collide with others, like a toothpick piercing the tree. To gain a feeling for how kinetic energy works, think about a speeding car versus a ball rolling slowly on the ground. The car, being much faster and heavier, has significantly more kinetic energy.
Work-Energy Principle
The work-energy principle is a handy tool in physics used to understand how forces interact with energy. This principle states that the work done by all the forces acting on an object equals the change in its kinetic energy. Mathematically, this can be expressed as:\[ W = \Delta KE \]Where \( W \) is work done and \( \Delta KE \) is the change in kinetic energy. In the case of the toothpick entering the branch, the work-energy principle tells us that the work done by the branch on the toothpick is equal to the toothpick's loss in kinetic energy. The kinetic energy initially present when the toothpick was moving is used to do work against the branch to penetrate it. By using the work-energy principle, we can determine how strong the force of the branch on the toothpick is.
Conversion of Units
Understanding how to convert units is vital for solving physics problems, as it ensures our calculations are consistent. For example, if you are working in meters and kilograms like in the international system of units (SI units), your computations become more straightforward and universally understandable. Let's look at two common conversions: grams to kilograms and millimeters to meters.
  • To convert grams into kilograms, divide by 1000, because 1000 grams make up a kilogram.
  • Similarly, to convert millimeters into meters, again divide by 1000, since there are 1000 millimeters in a meter.
In our exercise, we converted the mass of the toothpick from 0.13 grams to kilograms and the penetration depth from 15 millimeters to meters. These conversions helped us to apply formulas correctly and maintain precision during calculations. Always keep an eye on units; incorrect unit conversions can lead to huge errors!

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Most popular questions from this chapter

Tarzan, who weighs \(820 \mathrm{~N}\), swings from a cliff at the end of a \(20.0 \mathrm{~m}\) vine that hangs from a high tree limb and initially makes an angle of \(22.0^{\circ}\) with the vertical. Assume that an \(x\) axis extends horizontally away from the cliff edge and a \(y\) axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is \(760 \mathrm{~N}\). Just then, what are (a) the force on him from the vine in unit- vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis? What are the (e) magnitude and (f) angle of Tarzan's acceleration just then?

An \(80 \mathrm{~kg}\) man drops to a concrete patio from a window \(0.50 \mathrm{~m}\) above the patio. He neglects to bend his knees on landing, taking \(2.0 \mathrm{~cm}\) to stop. (a) What is his average acceleration from when his feet first touch the patio to when he stops? (b) What is the magnitude of the average stopping force exerted on him by the patio?

A lamp hangs vertically from a cord in a descending elevator that decelerates at \(2.4 \mathrm{~m} / \mathrm{s}^{2}\). (a) If the tension in the cord is \(89 \mathrm{~N}\), what is the lamp's mass? (b) What is the cord's tension when the elevator ascends with an upward acceleration of \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) ?

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thrust) of \(3260 \mathrm{~N}\), the craft descends at constant speed; if the engine provides only \(2200 \mathrm{~N}\), the craft accelerates downward at \(0.39\) \(\mathrm{m} / \mathrm{s}^{2}\). (a) What is the weight of the landing craft in the vicinity of Callisto's surface? (b) What is the mass of the craft? (c) What is the magnitude of the free-fall acceleration near the surface of Callisto?

A sphere of mass \(3.0 \times 10^{-4} \mathrm{~kg}\) is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle of \(37^{\circ}\) with the vertical. Find (a) the push magnitude and (b) the tension in the cord. *?0 (10) A dated box of dates, of mass \(5.00 \mathrm{~kg}\), is sent sliding up a frictionless ramp at an angle of \(\theta\) to the horizontal.
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