/*! 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 78 You and your bicycle have combin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 78

You and your bicycle have combined mass 80.0 \(\mathrm{kg} .\) When you reach the bridge, you are traveling along the road at 5.00 \(\mathrm{m} / \mathrm{s}(\) Fig. \(\mathrm{P} 6.78)\) . At the top of the bridge, you have climbed a vertical distance of 5.20 \(\mathrm{m}\) and have slowed to 1.50 \(\mathrm{m} / \mathrm{s} .\) You can ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

Short Answer

Expert verified
The total work done is 3162.8 J, and the cyclist does 3162.8 J of work.

Step by step solution

01

Understand the Problem

We need to determine the total work done on the bicycle system as the bike travels up a vertical incline while slowing down. This involves changes in both kinetic and potential energy.
02

Calculate Initial Kinetic Energy

The initial kinetic energy (KE) is calculated using the formula: \[ KE_i = \frac{1}{2} m v_i^2 \]where \( m = 80 \text{ kg} \) is the mass and \( v_i = 5.00 \text{ m/s} \) is the initial velocity.
03

Calculate Final Kinetic Energy

The final kinetic energy is given by: \[ KE_f = \frac{1}{2} m v_f^2 \]where \( v_f = 1.50 \text{ m/s} \) is the final velocity.
04

Calculate Change in Kinetic Energy

The change in kinetic energy (\( \Delta KE \)) is: \[ \Delta KE = KE_f - KE_i \]
05

Calculate Change in Potential Energy

The change in potential energy (\( \Delta PE \)) as the bike climbs is:\[ \Delta PE = mgh \]where \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity) and \( h = 5.20 \text{ m} \) is the height climbed.
06

Calculate Total Work Done

The total work done (\( W_{\text{total}} \)) is the sum of the change in kinetic and potential energies:\[ W_{\text{total}} = \Delta KE + \Delta PE \]
07

Calculate Work Done by the Cyclist

Since no non-conservative forces like friction are involved, the work done by the cyclist is equal to the total work done:\[ W_{\text{cyclist}} = W_{\text{total}} \]
08

Substitute Values and Solve

Substitute all known values and solve for \( W_{\text{total}} \) and \( W_{\text{cyclist}} \):\[ KE_i = \frac{1}{2} \times 80 \times (5.00)^2 = 1000 \text{ J} \]\[ KE_f = \frac{1}{2} \times 80 \times (1.50)^2 = 90 \text{ J} \]\[ \Delta KE = 90 - 1000 = -910 \text{ J} \]\[ \Delta PE = 80 \times 9.81 \times 5.20 = 4072.8 \text{ J} \]\[ W_{\text{total}} = -910 + 4072.8 = 3162.8 \text{ J} \]
09

Conclusion

The total work done on you and your bicycle is 3162.8 J, and the work you have done with the force you apply to the pedals is also 3162.8 J, as it's the only source of work due to external force.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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. It depends on two main factors: the mass of the object and its speed. The formula to calculate kinetic energy is \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.
  • An object with a larger mass or faster speed will have more kinetic energy.
  • Kinetic energy is a scalar quantity, which means it only has magnitude and no direction.
  • In the example of the cyclist, the initial kinetic energy was determined by the initial speed of 5.00 m/s and a mass of 80 kg, resulting in 1000 Joules of kinetic energy.
When the cyclist reaches to a speed of 1.50 m/s at the top of the bridge, the kinetic energy decreases to 90 Joules.
So, the change in kinetic energy as the cyclist travels up the bridge is \( -910 \text{ J} \), showing a loss as the speed decreases.
Potential Energy
Potential energy is the stored energy of position possessed by an object. It is closely related to an object's height in a gravitational field. The potential energy due to gravity can be calculated with the formula \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference point.
  • As an object is lifted higher, its potential energy increases.
  • Potential energy is also a scalar quantity and is affected by the position of the object in the gravitational field.
  • For the cycling example, as the cyclist ascends 5.20 m, the potential energy increases by 4072.8 Joules.
This gain in potential energy resulted from climbing, converting kinetic energy and work done by the cyclist into potential energy as they elevated to the top of the bridge.
Conservation of Energy
Conservation of energy is a fundamental principle stating that energy cannot be created or destroyed, only transformed from one form to another. In a closed system with no external forces acting on it, the total energy remains constant.
  • When analyzing the cyclist, we can see energy is converted between kinetic and potential energy as they climb the bridge.
  • The total work done by the cyclist, which accounts for energy transformation, is 3162.8 Joules.
  • Although individual energies change (kinetic decreases and potential increases), their sum鈥攖otal energy鈥攔emains constant in the absence of non-conservative forces like friction.
The work-energy theorem helps us understand these transformations clearly. It tells us that the work done on an object is equal to the change in its kinetic energy. Since no energy is lost to friction or external work in this scenario, we can conclude that all work exerted by the cyclist contributes to changing the kinetic and potential energies, satisfying the conservation of energy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

CP A 20.0 -kg crate sits at rest at the bottom of a 15.0 -m-long ramp that is inclined at \(34.0^{\circ}\) above the horizontal. A constant horizontal force of 290 \(\mathrm{N}\) is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 \(\mathrm{N}\) . (a) What is the total work done on the crate during its motion from the bottom to the top of the ramp? (b) How much time does it take the crate to travel to the top of the ramp?

You push your physics book 1.50 \(\mathrm{m}\) along a horizontal table top with a horizontal push of 2.40 \(\mathrm{N}\) while the opposing force of friction is 0.600 \(\mathrm{N} .\) How much work does each of the following forces do on the book: (a) your \(2.40-\mathrm{N}\) push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

A Near-Earth Asteroid. On April \(13,2029\) (Friday the 13th!), the asteroid 99942 Apophis will pass within \(18,600\) mi of the earth- about \(\frac{1}{13}\) the distance to the moon! It has a density of \(2600 \mathrm{kg} / \mathrm{m}^{3},\) can be modeled as a sphere 320 \(\mathrm{m}\) in diameter, and will be traveling at 12.6 \(\mathrm{km} / \mathrm{s}\) . (a) If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver? (b) The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases \(4.184 \times 10^{15}\) J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?

A 2.0 -kg box and a 3.0 -kg box on a perfectly smooth horizontal floor have a spring of force constant 250 \(\mathrm{N} / \mathrm{m}\) compressed between them. If the initial compression of the spring is \(6.0 \mathrm{cm},\) find the acceleration of each box the instant after they are released. Be sure to include free-body diagrams of each box as part of your solution.

Working Like a Horse. Your job is to lift 30 -kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. (a) How many crates would you have to load onto the truck in 1 minute for the average power output you use to lift the crates to equal 0.50 \(\mathrm{hp} ?\) (b) How many crates for an average power output of 100 \(\mathrm{W} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Astrophysics

Read Explanation

Space Physics

Read Explanation

Fluids

Read Explanation

Waves Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.