Consider the acceleration of a car on dry pavement, if there is no slipping. The axle moves at speed v, and the outside of the tire moves at a speed relative to the axle. The instantaneous velocity of the bottom of the tire is zero. How much work is done by the force exerted on the tire by the road? What is the source of the energy that increases the car’s translational kinetic energy?

Short Answer

Expert verified

The work done by the force exerted on the tire by the road is0 Joule.Theinternal energy of the car causespeeding a car and this will result ingaining translational kinetic energy.

Step by step solution

01

Identification of given data

  • The speed of a caroutside of theisVout=v
  • The instantaneous velocity of the bottom of the tire is Vout=0
02

Concept of the work done by the force

The work done by the force is determined by the multiplication of force and displacement.

03

Determination of the work done by the force

The work done by the force can be calculated as,

The motion of tires of a car is pure rolling. Therefore, the work done the road friction on the tires will be 0.Hence, the work done by the force exerted on the tire by the road is 0 Joule.

The engine of a car gives power to the wheel. The increase in the car's kinetic energy comes from the internal energy of the car, this will result in gaining translational kinetic energy.

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 Vaia!

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

Consider a system consisting of three particles:

m1=2kg,v1=(8,-6,15)m/sm2=6kg,v2=(-12,9,-6)m/sm3=4kg,v3=(-24,34,23)m/s

What isKrel, the kinetic energy of this system relative to the centre of mass?

If an object has a moment of inertia 19kg·m2 and rotates with an angular speed of 70rad/s, what is its rotational kinetic energy?

A uniform-density sphere whose mass is 10kgand radius is 0.4mmakes one complete rotation every0.2s. What is the rotational kinetic energy of the sphere?

You hold up an object that consists of two blocks at rest, each of massM=5kg, connected by a low-mass spring. Then you suddenly start applying a larger upward force of constant magnitudeF=167N(which is greater than2Mg). Figure9.60shows the situation some time later, when the blocks have moved upward, and the spring stretch has increased.

The heights of the centers of the two blocks are as follows:

Initial and final positions of block 1:y1i=0.3m,y1f=0.5m

Initial and final positions of block 2:y2i=0.7m,y2f=1.2m

It helps to show these heights on a diagram. Note that the initial center of mass of the two blocks isy1i+y1i/2, and the final center of mass of the two blocks isrole="math" localid="1656911769231" y1f+y1f/2. (a) Consider the point particle system corresponding to the two blocks and the spring. Calculate the increase in the total translational kinetic energy of the two blocks. It is important to draw a diagram showing all of the forces that are acting, and through what distance each force acts. (b) Consider the extended system corresponding to the two blocks and the spring. Calculate the increase of(Kvib+Us), the vibrational kinetic energy of the two blocks (their kinetic energy relative to the center of mass) plus the potential energy of the spring. It is important to draw a diagram showing all of the forces that are acting, and through what distance each force acts.

Tarzan, whose mass is 100kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.9m above the ground and the bottom of his dangling feet are at a height 2.1 above the ground. When he first hits the ground he has dropped a distance 2.1, so his center of mass is (2.9-2.1) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height above the ground. (a) Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground? (b) Consider the extended system. What is the net change in internal energy for Tarzan from just before his feet touch the ground to when he is in the crouched position?

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.

Sign-up for free