/*! 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 55 The (nonconservative) force prop... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The (nonconservative) force propelling a \(1.50 \times 10^{3}-\mathrm{kg}\) car up a mountain road does \(4.70 \times 10^{6} \mathrm{J}\) of work on the car. The car starts from rest at sea level and has a speed of \(27.0 \mathrm{m} / \mathrm{s}\) at an altitude of \(2.00 \times 10^{2} \mathrm{m}\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

Short Answer

Expert verified
The work done by friction and air resistance is \(-3.86 \times 10^{6} \ \mathrm{J}\).

Step by step solution

01

Identify Known Values

We are given the mass of the car as \( m = 1.50 \times 10^{3} \mathrm{kg} \), the initial velocity as \( v_i = 0 \ \mathrm{m/s} \) because the car starts from rest, the final velocity as \( v_f = 27.0 \ \mathrm{m/s} \), the altitude change as \( h = 2.00 \times 10^{2} \mathrm{m} \), and the work done by the propelling force as \( W_{propelling} = 4.70 \times 10^{6} \ \mathrm{J} \).
02

Calculate Change in Kinetic Energy

Kinetic energy is given by \( KE = \frac{1}{2} m v^2 \). The change in kinetic energy \( \Delta KE \) is \( \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \). Substituting the known values, \( \Delta KE = \frac{1}{2} (1.50 \times 10^{3}) (27.0)^2 - \frac{1}{2} (1.50 \times 10^{3}) (0)^2 \). Calculating this gives \( \Delta KE = 5.47 \times 10^{5} \ \mathrm{J} \).
03

Calculate Change in Potential Energy

Potential energy due to height is given by \( PE = mgh \). The change in potential energy \( \Delta PE \) is \( mgh \). Substituting the known values, \( \Delta PE = (1.50 \times 10^{3}) (9.8) (2.00 \times 10^{2}) \). Calculating this gives \( \Delta PE = 2.94 \times 10^{5} \ \mathrm{J} \).
04

Apply Work-Energy Principle

According to the work-energy principle, the work done by all forces \( W_{total} \) is equal to the change in kinetic energy plus the change in potential energy. So, \( W_{propelling} + W_{friction} = \Delta KE + \Delta PE \). Thus, \( W_{friction} = \Delta KE + \Delta PE - W_{propelling} \).
05

Solve for Work Done by Friction and Air Resistance

Using the equation from Step 4, plug in the values: \( W_{friction} = 5.47 \times 10^{5} + 2.94 \times 10^{5} - 4.70 \times 10^{6} \). Calculating this gives \( W_{friction} = -3.86 \times 10^{6} \ \mathrm{J} \). The negative sign indicates that the work done by friction and air resistance is opposite to the direction of motion.

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 91Ó°ÊÓ!

Key Concepts

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

Nonconservative Forces
Nonconservative forces are forces like friction, air resistance, or tension, which do not store mechanical energy in an isolated system and instead dissipate energy, usually as heat or sound. In the given exercise, the forces of friction and air resistance, both nonconservative, do work on the car moving uphill.
These forces are path-dependent, meaning the work done by nonconservative forces is influenced by the path taken, not just the initial and final positions, unlike conservative forces such as gravity.
Here’s why understanding nonconservative forces is vital:
  • They affect the total energy balance in a system. In the problem, they dissipated energy as the car traveled, which manifested as a negative work done of (\(-3.86 \times 10^{6} \, \mathrm{J}\)).
  • They must be considered when using the work-energy theorem, which states that the total work done is equal to the change in kinetic energy plus the change in potential energy.
Carefully considering the work done by nonconservative forces helps us understand more about the system's energy dynamics and how energy is conserved or transformed from one form to another.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is directly proportional to the mass of the object and the square of its velocity, defined by the equation \( KE = \frac{1}{2}mv^2 \).
In this exercise, the car's initial kinetic energy was zero because it started from rest. As it gained speed, reaching 27 m/s, its kinetic energy increased significantly.
The change in the car's kinetic energy was calculated to be \( \Delta KE = 5.47 \times 10^{5} \, \mathrm{J} \). This indicates the amount of energy the car gained due to its motion as it traveled uphill.
Understanding kinetic energy is essential as:
  • It highlights how much energy is needed to increase speed and how this energy is transformed from work or other energy forms.
  • It provides insight into the car's performance and efficiency under changing conditions.
Potential Energy
Potential energy refers to the stored energy of an object due to its position relative to other objects, particularly due to its height in the presence of a gravitational field. The formula \( PE = mgh \) calculates the potential energy, where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference level.
For the car in this problem, as it ascended to 200 m, it gained potential energy calculated as \( \Delta PE = 2.94 \times 10^{5} \mathrm{J} \).
This gain in potential energy, due to its elevated position, is crucial for:
  • Understanding the work-energy principle, where energy is transferred and converted but remains conserved in the system.
  • Evaluating how the car's energy transformations impact its ability to overcome gravitational forces while ascending.
Knowing how potential energy works helps us predict the energy requirements and changes for objects moving in gravitational fields, like cars climbing hills.

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

The drawing shows a skateboarder moving at \(5.4 \mathrm{m} / \mathrm{s}\) along a horizontal section of a track that is slanted upward by \(48^{\circ}\) above the horizontal at its end, which is \(0.40 \mathrm{m}\) above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height \(H\) to which she rises above the end of the track.

A person pulls a toboggan for a distance of \(35.0 \mathrm{m}\) along the snow with a rope directed \(25.0^{\circ}\) above the snow. The tension in the rope is \(94.0 \mathrm{N} .\) (a) How much work is done on the toboggan by the tension force? (b) How much work is done if the same tension is directed parallel to the snow?

Starting from rest, a \(1.9 \times 10^{-4}-\mathrm{kg}\) flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of \(0.38 \mathrm{N}\) on it. This force does \(+2.4 \times 10^{-4} \mathrm{J}\) of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.

In the sport of skeleton a participant jumps onto a sled (known as a skeleton) and proceeds to slide down an icy track, belly down and head first. In the 2010 Winter Olympics, the track had sixteen turns and dropped \(126 \mathrm{m}\) in elevation from top to bottom. (a) In the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed at the beginning of the run is relatively small and can be ignored. (b) In reality, the gold-medal winner (Canadian Jon Montgomery) reached the bottom in one heat with a speed of \(40.5 \mathrm{m} / \mathrm{s}\) (about \(91 \mathrm{mi} / \mathrm{h}\) ). How much work was done on him and his sled (assuming a total mass of \(118 \mathrm{kg}\) ) by nonconservative forces during this heat?

A bicyclist rides \(5.0 \mathrm{km}\) due east, while the resistive force from the air has a magnitude of \(3.0 \mathrm{N}\) and points due west. The rider then turns around and rides \(5.0 \mathrm{km}\) due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of \(3.0 \mathrm{N}\) and points due east. (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

See all solutions

Recommended explanations on Physics Textbooks

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.