/*! 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 94 The luxury liner Queen Elizabeth... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of \(92 \mathrm{MW}\) at a cruising speed of \(32.5\) knots. What forward force is exerted on the ship at this speed? \((1 \mathrm{knot}=1.852 \mathrm{~km} / \mathrm{h} .)\)

Short Answer

Expert verified
The forward force exerted is approximately 5.506 MN.

Step by step solution

01

Convert Speed to Meters per Second

First, we need to convert the ship's speed from knots to meters per second. The ship's speed is given as 32.5 knots.1 knot = 1.852 km/h, so:\[32.5 \text{ knots} \times 1.852 \frac{\text{kiometers}}{\text{hour}} = 60.14 \frac{\text{km}}{\text{hour}}\]Convert km/h to m/s using the conversion factor (1 km/h = 1/3.6 m/s):\[60.14 \frac{\text{km}}{\text{hour}} \times \frac{1}{3.6} \approx 16.71 \frac{\text{m}}{\text{s}}\].
02

Power Equation Understanding

The power exerted by the ship's engines at cruising speed is P = 92 MW. Power is the rate at which work is done (or energy is transferred), and it can be expressed as the product of force and velocity:\[P = F \times v\].Where P is power, F is force, and v is velocity.
03

Solve for Forward Force

Rearrange the power formula to solve for the forward force:\[F = \frac{P}{v}\].Substitute the given values, where P = 92 MW = 92 \times 10^6 W and v = 16.71 m/s:\[F = \frac{92 \times 10^6}{16.71} \approx 5.506 \times 10^6 \text{ N}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Power Conversion
Power conversion is an essential aspect in understanding how energy is transformed and utilized in different systems. In physics, power is the rate at which energy is transferred or converted. It is typically measured in watts (W) or megawatts (MW) in larger systems such as power plants or ships. One important formula to remember is that power can be calculated as the product of force and velocity:
\[ P = F \times v \]
  • Where \( P \) is power in watts, \( F \) is force in newtons, and \( v \) is velocity in meters per second.
Conversion of power units can often involve large numbers, especially when dealing with megawatts (1 MW = 1,000,000 W). Understanding this relationship helps in calculating other quantities like force when power and speed are known. This highlights the dynamic nature of power as not just a static quantity but as something related directly to force and motion dynamics.
Speed Conversion
Speed conversion is a fundamental skill in physics, as speeds are often given in different units and must be converted to a standard unit before any calculations can be performed. For the Queen Elizabeth 2, the speed was given in knots, a unit commonly used in maritime contexts, and needed to be converted to meters per second (m/s) for consistency in calculations.
Here's the step-by-step conversion process:
  • 1 knot equals 1.852 kilometers per hour (km/h).
  • Convert km/h to m/s using the factor: 1 km/h = \( \frac{1}{3.6} \) m/s.
For example, for a speed of 32.5 knots:
First convert knots to km/h:
\[ 32.5 \times 1.852 = 60.14 \text{ km/h} \]Then convert km/h to m/s:
\[ 60.14 \times \frac{1}{3.6} \approx 16.71 \text{ m/s} \]Using this method consistently ensures that all measurements are in standard units, making calculations more straightforward.
Force Calculation
Calculating force in scenarios involving motion and power requires a good grasp of how these quantities interrelate. From our power equation \( P = F \times v \), we rearrange it to
calculate force:
\[ F = \frac{P}{v} \]This means force can be found by dividing power by velocity. In the case of the Queen Elizabeth 2, where the power is provided as 92 MW and the speed is 16.71 m/s, converting the power to watts (since 1 MW = 1,000,000 W) gives:
\[ F = \frac{92 \times 10^6}{16.71} \]By solving this equation, you get the force the ship's engines exert, which turns out to be approximately:
\[ 5.506 \times 10^6 \text{ N} \]
  • This force is what propels the ship forward at the given speed of 32.5 knots.
  • Understanding this calculation shows the relationship between power, speed, and the resulting force exerted.
Energy Principles
The principles of energy are at the core of understanding mechanics and power systems. Energy can neither be created nor destroyed, only converted from one form to another. In the operation of the Queen Elizabeth 2, energy conversion is primarily from fuel energy in the engines to kinetic energy that propels the ship.
  • Kinetic energy, which is the energy of motion, is given by the equation: \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
But when discussing power, we're more focused on how rapidly this energy conversion occurs. Power, as mentioned, is the rate of doing work or converting energy – a reflection of how energy principles operate on a time basis.
Knowing these principles helps to understand why high-powered engines are needed for large ships and how this power is translated into the ship's movement through water. An appreciation of energy principles is crucial for optimizing and innovating solutions for efficient energy use in maritime technology.
  • These principles guide the engineering of systems to maximize performance while minimizing energy losses.
  • The emphasis is on converting as much engine power as possible into the forward motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

The string is \(L=120 \mathrm{~cm}\) long, has a ball attached to one end, and is fixed at its other end. A fixed peg is at point \(P\). Released from rest, the ball swings down until the string catches on the peg; then the ball swings up, around the peg. If the ball is to swing completely around the peg, what value must distance \(d \mathrm{ex}\) ceed? (Hint: The ball must still be moving at the top of its swing. Do you see why?)

During a rockslide, a \(520 \mathrm{~kg}\) rock slides from rest down a hillside that is \(500 \mathrm{~m}\) long and \(300 \mathrm{~m}\) high. The coefficient of kinetic friction between the rock and the hill surface is \(0.25 .(\mathrm{a})\) If the gravitational potential energy \(U\) of the rock-Earth system is zero at the bottom of the hill, what is the value of \(U\) just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

A \(3.2 \mathrm{~kg}\) sloth hangs \(3.0 \mathrm{~m}\) above the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point \(y=0\) to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

Shows a pendulum of length \(L=1.25 \mathrm{~m}\). Its bob (which effectively has all the mass) has speed \(v_{0}\) when the cord makes an angle \(\theta_{0}=40.0^{\circ}\) with the vertical. (a) What is the speed of the bob when it is in its lowest position if \(v_{0}=8.00 \mathrm{~m} / \mathrm{s} ?\) What is the least value that \(v_{0}\) can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? (d) Do the answers to (b) and (c) increase, decrease, or remain the same if \(\theta_{0}\) is increased by a few degrees?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.