/*! 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 11 You are testing a new amusement ... [FREE SOLUTION] | 91Ó°ÊÓ

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You are testing a new amusement park roller coaster with an empty car with mass 120 \(\mathrm{kg}\) . One part of the track is a vertical loop with radius 12.0 \(\mathrm{m}\) . At the bottom of the loop (point \(A\) ) the car has speed \(25.0 \mathrm{m} / \mathrm{s},\) and at the top of the loop (point \(B )\) it has speed 8.0 \(\mathrm{m} / \mathrm{s}\) . As the car rolls from point \(A\) to point \(B,\) how much work is done by friction?

Short Answer

Expert verified
The work done by friction is -5436 J.

Step by step solution

01

Calculate Initial and Final Kinetic Energy

The initial kinetic energy at the bottom (point A) can be calculated using the formula \( KE_i = \frac{1}{2} m v_i^2 \), where \( m \) is the mass and \( v_i \) is the initial velocity. Thus, \( KE_i = \frac{1}{2} \times 120 \, \mathrm{kg} \times (25.0 \, \mathrm{m/s})^2 = 37500 \, \mathrm{J} \). The final kinetic energy at the top (point B) is \( KE_f = \frac{1}{2} \times 120 \, \mathrm{kg} \times (8.0 \, \mathrm{m/s})^2 = 3840 \, \mathrm{J} \).
02

Determine Gravitational Potential Energy Gain

At the top of the loop (point B), the car also gains gravitational potential energy because it is 24 m above the bottom of the loop. This potential energy is calculated as \( PE_f = mgh \), where \( h = 24 \, \mathrm{m} \). So, \( PE_f = 120 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 24 \, \mathrm{m} = 28224 \, \mathrm{J} \).
03

Use Work-Energy Principle to Find Work Done by Friction

The work-energy principle states that the work done by non-conservative forces (such as friction) is equal to the change in mechanical energy (kinetic + potential energy). The initial mechanical energy is \( ME_i = KE_i = 37500 \, \mathrm{J} \). The final mechanical energy at point B is \( ME_f = KE_f + PE_f = 3840 \, \mathrm{J} + 28224 \, \mathrm{J} = 32064 \, \mathrm{J} \). Thus, the work done by friction is \( W_{friction} = ME_f - ME_i = 32064 \, \mathrm{J} - 37500 \, \mathrm{J} = -5436 \, \mathrm{J} \). This means 5436 J of energy is lost to friction.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. Imagine a roller coaster car speeding down a track – its kinetic energy increases because of its rapid movement. The formula for kinetic energy \[ KE = \frac{1}{2} mv^2 \]shows that kinetic energy depends on the mass of the object (\( m \)) and the square of its velocity (\( v \)). This relationship means that even a small increase in speed can result in a large increase in kinetic energy.

In the amusement park roller coaster problem, we calculated the car's initial kinetic energy at the bottom of the loop and its final kinetic energy at the top. At the bottom, the car had significantly higher kinetic energy because it was moving much faster. This energy change is central to understanding how the forces interact as the car travels through the loop.
Potential Energy
Potential energy is the stored energy an object obtains due to its position relative to other objects, particularly in a gravitational field. Think of it as the energy an object has "in reserve" that can be converted to other energy forms, like kinetic energy, when conditions change. A good example is a roller coaster at the top of a hill – all that height translates into stored energy just waiting to be released.

For gravitational potential energy specifically, the formula \[ PE = mgh \]helps us understand how mass (\( m \)), gravitational acceleration (\( g \)), and height (\( h \)) determine the potential energy of an object.

In our roller coaster scenario, as the car climbed from the bottom to the top of the loop, it gained potential energy because of its increased height. This gain was essential for the car's motion and energy balance as it completed the loop.
Friction
Friction is a force that opposes motion between two surfaces that are in contact. It acts in the opposite direction to movement, effectively "sucking" energy out of a system, which presents itself as work done on the objects moving against each other. This energy loss often results in heat generation.

In the roller coaster exercise, friction plays a key role by reducing the mechanical energy the car possesses. This loss is represented in the calculation of work done by friction: the difference between initial and final mechanical energies. When we calculated it, we found that 5436 J of energy was lost to friction. Understanding this concept highlights the importance of non-conservative forces in real-world scenarios, such as energy dissipation in entertainment rides.
Gravitational Force
Gravitational force is the attraction between two masses, like the Earth pulling objects towards its center. This force is why objects fall and why a roller coaster car stays on the track while going through a loop. Gravitational force impacts both potential and kinetic energy, guiding how the roller coaster car moves.

As the car in our exercise rises through the loop, gravitational force acts to increase its potential energy and decrease its kinetic energy, effectively slowing it down. At the same time, if the car descends, gravity helps convert this potential energy back to kinetic energy. The balance between gravitational force, kinetic energy, and potential energy is what keeps the roller coaster moving in such dramatic and thrilling ways.

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Most popular questions from this chapter

A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of 6.00 \(\mathrm{mup}\) the incline from \(A\) , the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

A 0.60\(\cdot \mathrm{kg}\) book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.2 \(\mathrm{N}\) . (a) How much work is done on the book by friction during a displacement of 3.0 \(\mathrm{m}\) to the left? (b) The book now slides 3.0 \(\mathrm{m}\) to the right, returning to its starting point. During this second 3.0 \(\mathrm{m}\) displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.

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A \(75-\mathrm{kg}\) roofer climbs a vertical 7.0 -m ladder to the flat roof of a house. He then walks 12 \(\mathrm{m}\) on the roof, climbs down another vertical \(7.0-\mathrm{m}\) ladder, and finally walks on the ground back to his starting point. How much work is done on him by gravity (a) as he climbs up; \((b)\) as he climbs down; (c) as he walks on the roof and on the ground? (d) What is the total work done on him by gravity during this round trip? On the basis of your answer to part (d), would you say that gravity is a conservative or nonconservative force? Explain.

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(550-\mathrm{N}\) kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

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