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A large cruise ship of mass \(6.50 \times 10^{7} \mathrm{~kg}\) has a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) at some instant. (a) What is the ship's kinetic cnergy at this time? (b) How much work is required to stop it? (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of \(2.50 \mathrm{~km}\) ?

Short Answer

Expert verified
The ship's kinetic energy is \(4.68 \times 10^{9} Joules\). The work required to stop it is also \(4.68 \times 10^{9} Joules\), and the constant force needed to stop it over a displacement of 2.50 km is \(1.872 \times 10^{6}\) Newtons.

Step by step solution

01

Compute the ship's kinetic energy

To calculate the ship's kinetic energy, use the formula \( KE = 1/2 m v^2 \). Substitute \( m = 6.50 \times 10^{7} kg \) and \( v = 12.0 m/s \) into the formula to get \( KE = 1/2 \times 6.50 \times 10^{7} kg \times (12.0 m/s)^2 = 4.68 \times 10^{9} Joules \)
02

Calculate the work required to stop the ship

The work required to stop the ship is equal to its kinetic energy. This is because work is defined as the energy transferred to or from an object via the application of force. Therefore, the work needed to stop the ship is \( W = KE = 4.68 \times 10^{9} Joules \)
03

Determine the constant force required to stop the ship

To find the force necessary to stop the ship, use the formula for work in terms of force and displacement \( W = Fd \), and rearrange for force \( F = W/d \). Now substitute the work \( W = 4.68 \times 10^{9} Joules \) and displacement \( d = 2.50 km = 2500 m \) into the formula to get \( F = 4.68 \times 10^{9} Joules / 2500 m = 1.872 \times 10^{6} Newtons \)

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Key Concepts

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

Work-Energy Theorem
Understanding the work-energy theorem is crucial when dealing with problems like the stopping of a massive cruise ship. Essentially, this theorem presents a relationship between the work done by all the forces acting on an object and the change in its kinetic energy.

The kinetic energy of an object is given by the formula \( KE = \frac{1}{2}mv^2 \), where \(m\) is the object’s mass and \(v\) its velocity. In our exercise, calculating the kinetic energy was the first step, providing us with a foundation for further calculations. As dictated by the work-energy theorem, the work (\(W\)) necessary to bring the ship to a stop is equivalent to the kinetic energy it possessed, since the ship's speed changes from 12.0 m/s to 0 m/s.

In this context, 'work' refers to the energy transfer needed to cause this change in motion. This is a prime example of the theorem in practice and sets the stage for understanding the rest of the problem.
Force and Motion
The interplay between force and motion lies at the heart of classical mechanics and, indeed, our cruise ship puzzle. Newton's second law of motion, which states that force is equal to mass times acceleration (\( F = ma \)), was subtly applied in our problem when we calculated the force necessary to stop the ship. However, we encountered the force in terms of work and displacement rather than acceleration.

The formula \( W = Fd \), where \( W \) is work, \( F \) is the average force applied, and \( d \) is the displacement, reveals that force can be considered in terms of energy and distance. The displacement in our problem is the distance over which the force is applied to bring the ship to a stop. By rearranging the work formula (\( F = \frac{W}{d} \) ), we found the magnitude of the constant force required.

This aspect of the exercise improves our understanding of how force is applied and exerted over distance to alter the motion of an object—in this case, to halt a massive ship.
Physics Problem Solving
Physics problem solving is a fundamental skill that is honed through practice and the understanding of core concepts, like the ones mentioned above. When addressing physics problems, it is essential to approach them systematically—breaking down complex scenarios into individual, more manageable concepts.

In the case of the cruise ship exercise, we followed a methodical approach. We first determined the ship's kinetic energy, then used the work-energy theorem to understand how much work would be required to stop it. Finally, we solved for the force needed over a specified distance. Each step built upon the previous, showcasing a structured approach to problem-solving in physics.

Cultivating these problem-solving skills involves identifying the known and unknown variables, translating words into equations, and carefully carrying out calculations. Through methodical practice, students can become adept at navigating the complexities of physics problem solving.

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Most popular questions from this chapter

Hooke's law describes a certain light spring of unstretched length \(35.0 \mathrm{~cm}\). When one end is attached to the top of a door frame and a \(7.50-\mathrm{kg}\) object is hung from the other end, the length of the spring is \(41.5 \mathrm{~cm}\). (a) Find its spring constant. (b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of \(190 \mathrm{~N}\). Find the length of the spring in this situation.

GP A \(60.0-\mathrm{kg}\) athlete leaps straight up into the air from a trampoline with an initial speed of \(9.0 \mathrm{~m} / \mathrm{s}\). The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is \(9.0 \mathrm{~m} / \mathrm{s}\) as \(y=0\). What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (c) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.

\(\mathrm{S}\) A projectile of mass \(m\) is fired horizontally with an initial speed of \(v_{0}\) from a height of \(h\) above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of \(m, v_{0}, h\), and \(g\) : (a) the work done by the force of gravity on the projectile, (b) the change in kinetic energy of the projectile since it was fired, and (c) the final kinetic energy of the projectile. (d) Are any of the answers changed if the initial angle is changed?

\(Q \mid C\) (a) A child slides down a water slide at an amusement park from an initial height \(h\). The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-ofenergy equation be modified? (e) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is \(12.0 \mathrm{~m}\).

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for the mechanical energy loss due to frictional forces exerted on the car by the air and the road. If the power developed by an engine is 175 hp, estimate the total frictional force acting on the car when it is moving at a speed of \(29 \mathrm{~m} / \mathrm{s}\). One horsepower cquals \(746 \mathrm{~W}\).

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