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\(\bullet\) \(\bullet\) The aircraft carrier \(U S S\) George Washington has mass \(1.0 \times 10^{8} \mathrm{kg} .\) When its engines are developing their full power of \(260,000\) hp, the George Washington travels at its top speed of 35 knots \((65 \mathrm{km} / \mathrm{h}) .\) If 70\(\%\) of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?

Short Answer

Expert verified
The force of water resistance opposing the carrier is approximately 7,518,544 N.

Step by step solution

01

Convert Power Units

First, we need to convert the power from horsepower to watts. We know that \[1 ext{ hp} = 746 ext{ watts}.\] Thus, the total power in watts is:\[260,000 ext{ hp} \times 746 ext{ watts/hp} = 193,960,000 ext{ watts}.\] Since only 70% of this power is applied to pushing the ship, the effective power is:\[0.70 \times 193,960,000 ext{ watts} = 135,772,000 ext{ watts}.\]
02

Convert Speed Units

Convert the speed of the aircraft carrier from knots to meters per second. We know:\[35 ext{ knots} = 65 ext{ km/h}.\]Convert kilometers per hour to meters per second using the conversion factor:\[ \frac{1 ext{ km/h}}{3.6} = ext{ m/s}.\]Thus, the speed is:\[\frac{65}{3.6} \approx 18.06 ext{ m/s}.\]
03

Calculate Force of Water Resistance

Using the relation between power, force, and velocity:\[P = F \cdot v,\]where \(P\) is the effective power and \(v\) is the velocity, we solve for \(F\) (the force of water resistance):\[F = \frac{P}{v} = \frac{135,772,000 ext{ watts}}{18.06 ext{ m/s}} \approx 7,518,544 ext{ N}.\] This is the magnitude of the force of water resistance that opposes the carrier's motion.

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Key Concepts

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

Power conversion
When dealing with Physics problems, understanding the conversion of power units can be crucial. Power, in this context, refers to the rate at which work is done or energy is transferred. It is often measured in watts in scientific applications but in horsepower (hp) in other contexts like automotive and marine situations. To convert horsepower to watts, remember the conversion factor:
  • 1 hp = 746 watts.
So, if we have a power of 260,000 hp, multiply it by 746 to get the power in watts.
That gives us:
  • 260,000 hp \( \times \) 746 W/hp = 193,960,000 watts.
Understanding this conversion allows us to apply power more effectively in calculations related to forces and movement.
Water resistance force
Water resistance, or drag force, is a critical factor when looking at objects moving through water, such as a ship. This force acts opposite to the direction of motion, essentially slowing the object down. It's vital to understand that the effective power applied by any marine engine has to overcome this resistance to maintain a certain speed.In our case, only 70% of the aircraft carrier's engine power is used to push it through the water. This "applied power" is what counters the water resistance. To evaluate water resistance:
  • Calculate the effective power available by multiplying total power by the percentage used: \( 0.70 \times 193,960,000 \text{ watts} = 135,772,000 \text{ watts} \).
The magnitude of water resistance can be determined with the relationship between power, force, and velocity, as it indicates how much force is needed to overcome the drag at a given speed.
Unit conversion
Unit conversion is a fundamental skill in Physics, allowing you to express measurements in a way that is most useful for solving problems. For speed, you'll often need to convert from one unit to another. Here, the speed of the aircraft carrier is given in both knots and kilometers per hour (km/h), but we usually need it in meters per second (m/s) for calculations involving force and power.Follow these steps for conversion:
  • 1 knot is approximately equal to 1.852 km/h.
  • The conversion from km/h to m/s is achieved via division by 3.6.
For the given speed:
  • 65 km/h \( \div \) 3.6 \( \approx \) 18.06 m/s.
Understanding unit conversions ensures you get the correct figures when doing calculations and helps maintain consistency across different measurement systems.
Force calculation
To figure out the force of water resistance, you rely on the fundamental equation that ties power, force, and velocity together: \( P = F \cdot v \). Here, \( P \) is the effective power in watts, \( F \) is the force in newtons, and \( v \) is the speed in meters per second.To solve for force:
  • Rearrange the formula to \( F = \frac{P}{v} \).
  • Insert the effective power and speed values.
Given:
  • \( P = 135,772,000 \text{ watts} \)
  • \( v = 18.06 \text{ m/s} \)
Calculate \( F \):
  • \( F = \frac{135,772,000}{18.06} \approx 7,518,544 \text{ newtons} \).
This force represents the power needed to overcome the resistance of water, highlighting the energy expenditure required for motion.
Speed conversion
Converting speed from one unit to another is a frequent task in Physics, especially in problems involving motion. Speed can be represented in many units, each suited for different types of analysis or applications.This exercise involves:
  • Converting knots to kilometers per hour and then to meters per second.
Initially, we know that 35 knots equals 65 km/h.
To proceed to meters per second, use the suitable conversion:
  • \( \frac{65}{3.6} \approx 18.06 \text{ m/s} \).
Correct speed conversion is vital as it ensures the calculations for other parameters like force are accurate. It gives you a common ground (m/s in this instance) to apply equations and derive solutions.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) A 12.0 N package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is supported from below by a vertical spring of force con- stant 325 \(\mathrm{N} / \mathrm{m}\) . If the pan has negligible weight, find the maxi- mum distance the spring will be compressed if no energy is dissipated by friction.

\(\bullet\) \(\bullet\) An unstretched spring has a force constant of 1200 \(\mathrm{N} / \mathrm{m}\) . How large a force and how much work are required to stretch the spring: (a) by 1.0 \(\mathrm{m}\) from its unstretched length, and (b) by 1.0 \(\mathrm{m}\) beyond the length reached in part (a)?

\(\bullet\) \(\bullet\) A 250 g object on a frictionless, horizontal lab table is pushed against a spring of force constant 35 \(\mathrm{N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is com- pressed 12.0 \(\mathrm{cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

\(\bullet\) To stretch a certain spring by 2.5 \(\mathrm{cm}\) from its equilibrium position requires 8.0 \(\mathrm{J}\) of work. (a) What is the force constant of this spring? (b) What was the maximum force required to stretch it by that distance?

\(\cdot\) \(\cdot\) Two tugboats pull a disabled supertanker. Each tug exerts a constant force of \(1.80 \times 10^{6} \mathrm{N}\) , one \(14^{\circ}\) west of north and the other \(14^{\circ}\) east of north, as they pull the tanker 0.75 \(\mathrm{km}\) toward the north. What is the total work they do on the supertanker?

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