/*! 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 53 The Great Sandini is a \(60-\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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The Great Sandini is a \(60-\mathrm{kg}\) circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 \(\mathrm{N} / \mathrm{m}\) that he will compress with a force of 4400 \(\mathrm{N}\) . The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40 \(\mathrm{N}\) during the 4.0 \(\mathrm{m}\) he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 \(\mathrm{m}\) above his initial rest position?

Short Answer

Expert verified
15.5 m/s

Step by step solution

01

Calculate the compression of the spring

The force used to compress the spring is equal to the spring constant multiplied by the distance compressed, given by the formula: \[ F = kx \]Rearranging for compression distance \( x \), we have: \[ x = \frac{F}{k} = \frac{4400 \, \text{N}}{1100 \, \text{N/m}} = 4 \, \text{m} \]
02

Compute the elastic potential energy stored in the spring

The elastic potential energy (EPE) stored in the spring is given by: \[ \text{EPE} = \frac{1}{2} k x^2 = \frac{1}{2} \times 1100 \, \text{N/m} \times (4 \, \text{m})^2 = 8800 \, \text{J} \]
03

Calculate the work done by friction

The work done against friction is given by the force of friction multiplied by the distance: \[ W_{friction} = f \cdot d = 40 \, \text{N} \times 4 \, \text{m} = 160 \, \text{J} \]
04

Determine gravitational potential energy change

The gravitational potential energy change (GPE) as the performer moves 2.5 m higher is: \[ \Delta \text{GPE} = mgh = 60 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 2.5 \, \text{m} = 1471.5 \, \text{J} \]
05

Use conservation of energy to find final kinetic energy

The initial elastic potential energy is converted into kinetic energy (KE), with some energy lost to friction and increased potential energy. Set up the equation:\[ \text{EPE} - W_{friction} - \Delta \text{GPE} = \text{KE} \]\[ 8800 \, \text{J} - 160 \, \text{J} - 1471.5 \, \text{J} = \text{KE} \]\[ \text{KE} = 7168.5 \, \text{J} \]
06

Solve for the final velocity

The kinetic energy is related to speed by the formula:\[ \text{KE} = \frac{1}{2} mv^2 \]Solving for \( v \):\[ v = \sqrt{\frac{2 \times 7168.5 \, \text{J}}{60 \, \text{kg}}} \approx 15.5 \, \text{m/s} \]

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Key Concepts

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

Energy Conservation
Energy conservation is a vital principle in physics which states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of our exercise, the initial elastic potential energy stored in the compressed spring is transferred into The Great Sandini’s kinetic energy as he moves through the barrel.
However, some energy is also transferred to overcome friction and to increase his gravitational potential energy as he rises by 2.5 meters.
The conservation equation used is:
  • Initial Elastic Potential Energy = Final Kinetic Energy + Work done against friction + Change in Gravitational Potential Energy
Understanding this equation allows us to determine the speed at which The Great Sandini emerges from the cannon by accounting for all forms of energy transformation involved in his motion.
Kinetic Energy
Kinetic energy is the energy of motion. It's defined by the formula:
  • \( \text{KE} = \frac{1}{2} mv^2 \)
where \( m \) is the mass and \( v \) is the velocity. In this exercise, once The Great Sandini exits the spring gun, his kinetic energy is the leftover energy after losses to friction and potential energy changes.
To compute his final speed, we rearrange the formula for kinetic energy to solve for velocity, indicating how fast he travels once all energy transfers are accounted for.
Potential Energy
Potential energy is stored energy that an object has due to its position or state. The primary forms considered here are elastic potential energy, from the compressed spring, and gravitational potential energy, due to The Great Sandini's elevated position.
Gravitational potential energy is given by:
  • \( \Delta \text{GPE} = mgh \)
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height difference. As he rises to a height of 2.5 m, some of the initial energy converts into gravitational potential, reducing the kinetic energy he has available.
Elastic Potential Energy
Elastic potential energy is energy stored in objects that can be stretched or compressed, like springs. The energy stored in the spring is determined by the formula:
  • \( ext{EPE} = \frac{1}{2} k x^2 \)
where \( k \) is the spring constant, and \( x \) is the compression distance of the spring.
In the problem, initial compression of the spring captures energy ready to be converted into kinetic motion.
This form of energy is crucial in powering The Great Sandini's launch and must overcome all energy losses to define his speed.
Work Against Friction
Work against friction is the energy required to overcome resistive forces that oppose motion. In this context, it refers to the force needed to move The Great Sandini against the friction within the cannon barrel coated with Teflon.
This work is calculated by:
  • \( W_{friction} = f \cdot d \)
where \( f \) is the frictional force and \( d \) is the distance moved. Understanding how friction reduces the total energy available for The Great Sandini's motion helps us appreciate its role in real-world energy conversions.
In our exercise, work done against friction decreases the kinetic energy that could otherwise be used for velocity, thus impacting his speed.

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Most popular questions from this chapter

Bouncing Ball. A 650 -gram rubber ball is dropped from an initial height of \(250 \mathrm{m},\) and on each bounce it returns to 75\(\%\) of its previous height. (a) What is the initial mechanical energy of the ball, just after it is released from its initial height? (b) How much mechanical energy does the ball lose during its first bounce? What happens to this energy? (c) How much mechanical energy is lost during the second bounce?

A \(60.0-\mathrm{kg}\) skier starts from rest at the top of a ski slope 65.0 \(\mathrm{m}\) high. (a) If frictional forces do \(-10.5 \mathbf{k J}\) of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow, where \(\mu_{\mathrm{k}}=0.20\) If the patch is 82.0 \(\mathrm{m}\) wide and the average force of air resistance on the skier is 160 \(\mathrm{N}\) , how fast is she going after crossing the patch? (c) The skier hits a snowdrift and Penetrates 2.5 \(\mathrm{m}\) into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the middle of the gym floor, as shown in Fig. \(7.26 .\) You take your physics book and push it from one person to the other. The book has a mass of 1.5 \(\mathrm{kg}\) , and the coefficient of kinetic friction between the book and the floor is \(\mu_{\mathrm{x}}=0.25 .\) (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connecting these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim who then slides it back to you. What is the total work done by friction during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

A \(0.100-\mathrm{kg}\) potato is tied to a string with length \(2.50 \mathrm{m},\) and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

A 2.50 -kg mass is pushed against a horizontal spring of force constant 25.0 \(\mathrm{N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 J of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

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