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\(\bullet\) The engine of a motorboat delivers 30.0 \(\mathrm{kW}\) to the propeller while the boat is moving at 15.0 \(\mathrm{m} / \mathrm{s} .\) What would be the ten- sion in the towline if the boat were being towed at the same speed?

Short Answer

Expert verified
Tension in the towline is 2,000 N.

Step by step solution

01

Understanding Power and Force

The power delivered by the engine to the propeller can also be expressed as the product of force (tension in the towline) and speed. Since the engine power is known to be 30.0 kW and the speed of the boat is 15.0 m/s, we can start with the relationship between power, force, and velocity.
02

Power Equation

The formula for power in terms of force and velocity is given by: \[ P = F \cdot v \] where \( P \) is power (30.0 kW), \( F \) is force, and \( v \) is velocity (15.0 m/s). Here, \( F \) is the tension we need to find.
03

Convert Power Units

First, convert the power from kilowatts to watts because the standard unit of power is watts. \[ 30.0 \text{ kW} = 30,000 \text{ W} \]
04

Solve for Force

Rearrange the power equation to solve for force:\[ F = \frac{P}{v} \]Substitute the given values:\[ F = \frac{30,000 \, \text{W}}{15.0 \, \text{m/s}} = 2,000 \, \text{N} \]Thus, the tension in the towline would be 2,000 N.

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Key Concepts

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

Power and energy
Power and energy are fundamental concepts in physics used to describe how work is done and the rate at which work is done, respectively. In our exercise, the focus is on power, which is the ability of the engine to do work over a given time. The motorboat engine generates 30.0 kW of power.
Power is quantified as energy transfer per unit time, and is expressed in watts (W), where 1 W is equal to 1 joule per second. The boat's engine delivers power to the propeller, enabling it to move at a constant speed. Here, power links to both force and velocity, bridging the two concepts using the formula:
  • \( P = F \cdot v \)
This equation helps us understand how much force (in terms of tension) is needed when power is given, highlighting the energy's constant flow to keep the boat moving steadily.
Force and tension
In physics, force is an interaction that causes a mass to accelerate, while tension refers to a specific type of force transmitted through a string, cable, or similar object. In this exercise, the tension represents the force that would be needed to tow the boat.
When the engine delivers power to the propeller, it generates a forward force that enables the boat's movement.   This forward force is essentially the tension in the scenario where the boat is being towed at the same speed. We determine the tension through the relationship between power and velocity. This requires reworking the power equation to establish how much force is involved:
  • \( F = \frac{P}{v} \)
This formula tells us the tension in terms of the known power and speed values.
Unit conversion
Unit conversion is crucial in physics to ensure that all quantities in an equation are in compatible units. In this specific problem, the power supplied by the engine needs to be converted from kilowatts to watts. This conversion is necessary because the standard metric unit of power is the watt.
1 kilowatt (kW) is equivalent to 1,000 watts (W). Therefore, 30.0 kW must be converted to 30,000 W to facilitate straightforward calculations in the formula:
  • 30.0 kW = 30,000 W
By ensuring that calculations are conducted in standard units, we maintain consistency and accuracy, enabling us to reliably solve the problem.
Proper unit conversion prevents errors and eliminates any misinterpretation of results, especially when engaging with complex physics equations.
Physics equations
Physics equations are mathematical representations that describe relationships between different physical quantities. In our problem, we use the power equation to connect power output, force, and velocity.
The primary equation utilized is:
  • \( P = F \cdot v \)
By rearranging this equation into \( F = \frac{P}{v} \), we can solve for force when power and velocity are known. This manipulation is a common technique in physics to isolate an unknown variable.
Such equations help bridge theoretical physics concepts with practical, real-world applications, allowing predictions of how systems will behave under specific conditions.
Understanding these principles and knowing how to apply them enables effective problem-solving across various physical scenarios.

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Most popular questions from this chapter

\(\bullet\) How high can we jump? The maximum height a typical human can jump from a crouched start is about 60 \(\mathrm{cm} .\) By how much does the gravitational potential energy increase for a 72 \(\mathrm{kg}\) person in such a jump? Where does this energy come from?

\(\bullet\) \(\bullet\) \(\mathrm{A} 25 \mathrm{kg}\) child plays on a swing having support ropes that are 2.20 \(\mathrm{m}\) long. A friend pulls her back until the ropes are \(42^{\circ}\) from the vertical and releases her from rest. (a) What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? (b) How fast will she be moving at the bottom of the swing? (c) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

\(\cdot\) A fisherman reels in 12.0 \(\mathrm{m}\) of line while landing a fish, using a constant forward pull of 25.0 \(\mathrm{N}\) . How much work does the tension in the line do on the fish?

\(\bullet\) \(\bullet\) Maximum sustainable human power. The maximum sustainable mechanical power a human can produce is about \(\frac{1}{3}\) hp. How many food calories can a human burn up in an hour by exercising at this rate? (Remember that only 20\(\%\) of the food energy used goes into mechanical energy.)

\(\cdot\) A boat with a horizontal tow rope pulls a water skier. She skis off to the side, so the rope makes an angle of \(15.0^{\circ}\) with the forward direction of motion. If the tension in the rope is \(180 \mathrm{N},\) how much work does the rope do on the skier during a forward displacement of 300.0 \(\mathrm{m} ?\)

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