/*! 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 4 A catapult has two rubber arms, ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 4

A catapult has two rubber arms, each with a square cross-section with a width \(4 \mathrm{~mm}\) and length \(300 \mathrm{~mm}\). In use its arms are stretched to three times their original length before release. Assume the modulus of rubber is \(10^{-3} \mathrm{GPa}\) and that it does not change when the rubber is stretched. How much energy is stored in the catapult just before release?

Short Answer

Expert verified
The energy stored in the catapult just before release is 19.2 J.

Step by step solution

01

- Understand the Given Data

The problem provides the following information: - Width of the rubber arm: 4 mm - Original length of the rubber arm: 300 mm - Stretched length of the rubber arm: 3 times the original length - Modulus of rubber: \(10^{-3} \text{ GPa}\)
02

- Convert Units

Convert all measurements to consistent units (preferably meters and Pascals). - Width: 4 mm = 0.004 m - Original length: 300 mm = 0.3 m - Stretched length: \(3 \times 0.3 \text{ m} = 0.9 \text{ m}\) - Modulus of rubber: \(10^{-3} \text{ GPa} = 10^{-3} \times 10^9 \text{ Pa} = 10^6 \text{ Pa}\)
03

- Calculate the Strain

Strain, which is the ratio of the change in length to the original length, is given by: \[ \text{Strain} = \frac{\text{Stretched Length} - \text{Original Length}}{\text{Original Length}} = \frac{0.9 - 0.3}{0.3} = 2 \text{ (or 200%)} \]
04

- Calculate the Stress

Stress can be calculated using the modulus of elasticity (Young's modulus) and strain: \[ \text{Stress} = \text{Modulus} \times \text{Strain} = 10^6 \text{ Pa} \times 2 = 2 \times 10^6 \text{ Pa} \]
05

- Calculate the Cross-Sectional Area

The cross-sectional area of the arm (which is square) is given by: \[ \text{Area} = (\text{Width})^2 = (0.004 \text{ m})^2 = 1.6 \times 10^{-5} \text{ m}^2 \]
06

- Calculate Force

Force can be calculated using the formula: \[ \text{Force} = \text{Stress} \times \text{Area} = 2 \times 10^6 \text{ Pa} \times 1.6 \times 10^{-5} \text{ m}^2 = 32 \text{ N} \]
07

- Calculate Work Done (Energy Stored)

The work done (energy stored) in stretching the rubber arm is given by the formula for elastic potential energy: \[ \text{Energy} = \frac{1}{2} \times \text{Force} \times \text{extension} = \frac{1}{2} \times 32 \text{ N} \times (0.9 \text{ m} - 0.3 \text{ m}) = \frac{1}{2} \times 32 \times 0.6 = 9.6 \text{ J} \] Since there are two arms, the total energy stored is: \[ 9.6 \times 2 = 19.2 \text{ J} \]

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Key Concepts

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

elastic potential energy
Elastic potential energy is the energy stored in elastic materials when they are stretched or compressed. In this problem, the catapult鈥檚 rubber arms store this energy when stretched. The formula to find elastic potential energy is: In the given exercise, we calculated the energy stored in each rubber arm to be 9.6 J using the formula and multiplying this by 2 for both arms.
Young's modulus
Young's modulus is a fundamental property of elastic materials that measures their stiffness. It is defined as the ratio of stress to strain in a material under elastic deformation: Young's modulus is crucial because it tells us how much a material will stretch or compress under a given force without permanently deforming. When rubber is stretched in our exercise, its modulus remains constant at 10^6 Pa.
strain and stress
Strain and stress are two key concepts in material deformation: - **Strain**: It measures how much deformation occurs in a material compared to its original size, calculated as the ratio of change in length to original length: When the rubber arm stretches from 0.3 m to 0.9 m, the strain is 2 (200%). - **Stress**: It measures the internal force per unit area within a material and is calculated using the modulus of elasticity and strain: In our exercise, stress is given by: For the rubber arm, the stress calculation is 2 x 10^6 Pa.
cross-sectional area calculation
The cross-sectional area is vital for determining how a force applied across a material鈥檚 surface distributes internally. For a square cross-section, the area is calculated using: In the exercise, each rubber arm has a width of 0.004 m, making its cross-sectional area:
unit conversion in physics
Unit conversion ensures consistency and accuracy when solving physics problems. Common conversions include: - Millimeters to meters (1 mm = 0.001 m) - Gigapascals to Pascals (1 GPa = 10^9 Pa) The problem converts the rubber arm width from 4 mm to 0.004 m and the modulus from 10^{-3} GPa to 10^6 Pa.

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Most popular questions from this chapter

The cable of a hoist has a cross-section of \(80 \mathrm{~mm}^{2}\). The hoist is used to lift a crate weighing \(500 \mathrm{~kg}\). What is the stress in the cable? The free length of the cable is \(3 \mathrm{~m}\). How much will it extend if it is made of steel (modulus \(200 \mathrm{GPa}\) )? How much if it is made of polypropylene, PP (modulus \(1.2 \mathrm{GPa}\) )?

What is meant by: \- A crystalline solid? \- An amorphous solid? \- A thermoplastic? \- A thermoset? \- An elastomer?
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