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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 tension in the towline if the boat were being towed at the same speed?

Short Answer

Expert verified
The tension in the towline is 2,000 N.

Step by step solution

01

Understand the Relationship Between Power and Force

The power delivered by the engine, which is measured in watts or kilowatts, is the rate at which work is done. The power is defined as the product of the force exerted in the direction of the movement and the velocity of the object. The formula for power is given by \( P = F \times v \), where \( P \) is power in watts, \( F \) is the force in newtons, and \( v \) is the velocity in meters per second.
02

Identify Known Values

We are given the power output \( P = 30.0 \, \mathrm{kW} = 30,000 \, \mathrm{W} \) and the speed \( v = 15.0 \, \mathrm{m/s} \). These values will be used in the equation \( P = F \times v \) to solve for the force \( F \), which is the tension in the towline.
03

Rearrange the Power Equation to Solve for Force

We need to find the tension (force), so we rearrange the power equation to solve for \( F \):\[ F = \frac{P}{v} \]Substitute the known values of power and velocity into this equation.
04

Substitute and Calculate

Substitute \( P = 30,000 \, \mathrm{W} \) and \( v = 15.0 \, \mathrm{m/s} \) into the rearranged equation:\[ F = \frac{30,000}{15.0} \]Now, perform the division to calculate the force.
05

Finalize the Calculation

Carry out the division to find the force:\[ F = 2,000 \, \mathrm{N} \]Thus, the tension in the towline would be \( 2,000 \, \mathrm{N} \).

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Key Concepts

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

Power and Force Relationship
Power and force are key concepts in physics that describe how energy is transferred and how objects move. The relationship between power and force is essential to understand when solving problems involving moving objects. Power is defined as the rate at which work is done or energy is transferred within a system. It is measured in watts (W) or kilowatts (kW).

Force, on the other hand, is a vector quantity that represents the interaction capable of changing an object's motion. It is measured in newtons (N). The equation that links power, force, and velocity is given by:
  • Power, \( P = F \times v \)
where:
  • \( P \) is the power in watts,
  • \( F \) is the force in newtons,
  • \( v \) is the velocity in meters per second.
This equation indicates that the power used to move an object is directly proportional to the product of the force applied and the velocity of movement. By understanding this relationship, you can rearrange the equation to solve for any unknown variable.
Tension in Towline
Tension refers to the force transmitted through a string, cable, or, in this case, a towline. When determining the tension in a towline, we are essentially calculating the force needed to tow an object at a constant speed.

In the problem, the motorboat's engine delivers a specific amount of power while moving at a constant velocity. In this scenario of constant velocity, the tension in the towline will be equal to the force required by the engine to maintain that speed. We use the equation:
  • \( F = \frac{P}{v} \)
By inputting the power output of the engine and the boat's velocity, you can determine the tension in the towline. This relationship highlights the efficiency of energy transfer from the engine to the movement of the boat at the given speed.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause such motion. It involves parameters like displacement, velocity, and acceleration. In this exercise, while direct kinematic equations are not used, understanding the concept of constant velocity is crucial.

This concept implies that the boat’s velocity remains the same as it is being towed, signifying a balanced force scenario where the net force acting on the boat is zero. Fundamental kinematic principles state that when an object moves at a constant velocity, the sum of the forces acting upon it is zero, allowing us to focus on calculating the required force or tension to maintain that velocity.
Work and Power Calculations
Work and power calculations form the foundation for understanding how energy is utilized in mechanical systems. Work involves the process of energy being transferred through the movement of an object via a force. It is calculated as the product of the force and the distance moved in the direction of the force.

Power, however, focuses on the time rate of doing work and is a measure of how much work is completed over a period.
  • Work, \( W = F \times d \)
  • Power, \( P = \frac{W}{t} \)
where:
  • \( W \) is the work done in joules,
  • \( d \) is the distance in meters,
  • \( t \) is the time in seconds.
For this exercise, we need to consider the power equation, \( P = F \times v \), to calculate the force required to tow the boat at a steady speed efficiently. Understanding these concepts equips students with the necessary tools to tackle a variety of practical physics problems.

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Most popular questions from this chapter

Your friend (mass \(65.0 \mathrm{~kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for \(3.00 \mathrm{~s}\) and accelerate your friend from rest to a speed of \(6.00 \mathrm{~m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

A fun-loving \(11.4 \mathrm{~kg}\) otter slides up a hill and then back down to the same place. If she starts up at \(5.75 \mathrm{~m} / \mathrm{s}\) and returns at \(3.75 \mathrm{~m} / \mathrm{s}\) how much mechanical energy did she lose on the hill, and what happened to that energy?

A ball is thrown upward with an initial velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

A good workout. You overindulged in a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide to do a series of arm raises while holding a \(5.0 \mathrm{~kg}\) weight in one hand. The distance from your elbow to the weight is \(35 \mathrm{~cm}\), and in each arm raise you start with your arm horizontal and pivot it until it is vertical. Assume that the weight of your arm is small enough compared with the weight you are lifting that you can ignore it. As is typical, your muscles are \(20 \%\) efficient in converting the food energy into mechanical energy, with the rest going into heat. If your dessert contained 350 food calories, how many arm raises must you do to work off these calories? Is it realistic to do them all in one session?

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?

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