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Chapter 7: Problem 53

The Great Sandini is a 60 -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 \(\mathrm{N} / \mathrm{m} / \mathrm{m}\) 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
The speed at which he emerges is approximately 15.5 m/s.

Step by step solution

01

Calculate Spring Compression

To find how much the spring is compressed, use Hooke's Law, which states: \[ F = k imes x \]where \( F \) is the applied force of 4400 N and \( k \) is the spring constant of 1100 N/m. Solving for \( x \):\[ x = \frac{F}{k} = \frac{4400}{1100} = 4 \text{ m} \]
02

Compute Work Done by Spring

The work done by the spring as it releases is calculated by the formula for elastic potential energy:\[ W_s = \frac{1}{2} k x^2 \]Substitute \( k = 1100 \text{ N/m} \) and \( x = 4 \text{ m} \) to get:\[ W_s = \frac{1}{2} \times 1100 \times (4)^2 = 8800 \text{ J} \]
03

Calculate Work Done Against Friction

The work done against friction is given by:\[ W_f = f \times d \]where \( f = 40 \text{ N} \) is the friction force, and \( d = 4 \text{ m} \) is the distance moved in the barrel:\[ W_f = 40 \times 4 = 160 \text{ J} \]
04

Determine Kinetic Energy at Barrel Exit

The kinetic energy at the barrel's exit is the net work done, which is the work done by the spring minus the work done against friction:\[ KE = W_s - W_f = 8800 - 160 = 8640 \text{ J} \]
05

Account for Gravitational Potential Energy Change

The Great Sandini is 2.5 m above the initial position, so we need to account for the change in gravitational potential energy:\[ \Delta PE = mgh \]where \( m = 60 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 2.5 \text{ m} \):\[ \Delta PE = 60 \times 9.8 \times 2.5 = 1470 \text{ J} \]
06

Calculate Final Speed

The net kinetic energy at the exit includes the decrease due to gravitational potential energy:\[ KE_{final} = KE - \Delta PE = 8640 - 1470 = 7170 \text{ J} \]Using the kinetic energy formula, \( KE = \frac{1}{2} mv^2 \), solve for \( v \):\[ 7170 = \frac{1}{2} \times 60 \times v^2 \]\[ 7170 = 30v^2 \]\[ v^2 = \frac{7170}{30} = 239 \]\[ v = \sqrt{239} \approx 15.5 \text{ m/s} \]

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Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental concept that explains how springs work. It relates the force applied to a spring to the distance it compresses or stretches. The formula is \[ F = k \times x \] where
  • \( F \) is the force applied,
  • \( k \) is the spring constant, a measure of the spring's stiffness,
  • \( x \) is the displacement from its rest position.
In our exercise, a 4400 N force compresses a spring with a constant of 1100 N/m. By rearranging the formula and calculating, we determine that the spring compresses by 4 meters.
This law helps us understand the initial potential energy stored in the spring, crucial for calculating further dynamics like kinetic energy.
Kinetic Energy
Kinetic energy is the energy of motion. When The Great Sandini is shot from the cannon, his kinetic energy increases as he speeds through the barrel. The equation for kinetic energy is \[ KE = \frac{1}{2} mv^2 \] where
  • \( m \) is the mass of the object,
  • \( v \) is the velocity.
Initially, the work done by the spring鈥攁ll of the energy stored鈥攃onverts into kinetic energy as Sandini exits the barrel. However, friction also plays a role.
The kinetic energy acquired at the end of the barrel is slightly reduced by work done against friction, allowing us to find the speed of Sandini when he leaves the barrel. Kinetic energy is essential for figuring out the speed and energy dynamics of objects in motion.
Gravitational Potential Energy
Gravitational potential energy (GPE) comes into play when there are changes in height. As The Great Sandini rises by 2.5 meters, the potential energy due to gravity changes. The formula to calculate this energy is \[ \Delta PE = mgh \] where
  • \( m \) is the mass,
  • \( g \) is the acceleration due to gravity, typically 9.8 m/s extsuperscript{2},
  • \( h \) is the height change.
In our case, the potential energy increase is calculated to be 1470 Joules as Sandini moves upward.
Understanding GPE helps explain how energy is distributed between kinetic and potential forms and ensures conservation within the system.
Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in kinetic energy. This principle helps us track energy transformations through different forms, such as potential and kinetic energy, and provides a comprehensive view of energy conservation.
For Sandini, the spring's work minus the friction's work gives the net kinetic energy at the barrel's end. Additionally, this kinetic energy changes as gravitational potential energy increases when rising.
  • Spring work becomes kinetic energy.
  • Friction reduces this energy.
  • Rising height transfers some kinetic energy into gravitational potential energy.
The work-energy principle helps ensure energy conservation across different forces and motions. It confirms that while forms may change, the total energy conserved leads to predictive outcomes for speed and position.

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Most popular questions from this chapter

Cp A small block with mass 0.0500 kg slides in a vertical circle of radius \(R=0.800 \mathrm{m}\) on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 \(\mathrm{N} .\) What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

A force of 800 \(\mathrm{N}\) stretches a certain spring a distance of 0.200 \(\mathrm{m}\) . (a) What is the potential energy of the spring when it is stretched 0.200 \(\mathrm{m} ?\) (b) What is its potential energy when it is compressed 5.00 \(\mathrm{cm} ?\)

A \(0.150-\mathrm{kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 \(\mathrm{m}\) above the floor. The spring has force constant 1900 \(\mathrm{N} / \mathrm{m}\) and is initially compressed 0.045 \(\mathrm{m} .\) The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

Bi0 Food Calories. The food calorie, equal to 4186 \(\mathrm{J}\) . is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 140 food calories per bar. (a) If a \(65-\mathrm{kg}\) hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy? (b) If, as is typical, only 20\(\%\) of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that 100\(\%\) of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)

A 10.0 -kg box is pulled by a horizontal wire in a circle on a rough horizontal surface for which the coefficient of kinetic friction is \(0.250 .\) Calculate the work done by friction during one complete circular trip if the radius is (a) 2.00 \(\mathrm{m}\) and (b) 4.00 \(\mathrm{m}\) . (c) On the basis of the results you just obtained, would you say that friction is a conservative or nonconservative force? Explain.
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