Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
Question
Book Icon
Chapter 2, Problem 2.14P

(a)

To determine

The velocity of the mass with respect to time.

(a)

Expert Solution
Check Mark

Answer to Problem 2.14P

The velocity of the mass with respect to time is v(t)=vln(F0tmv+ev0/v).

Explanation of Solution

Given, the force on the mass m is F=F0ev'/v

Write the expression for t from problem 2.7

    t=mv0vdv'F(v')        (I)

Substituting F=F0ev'/v in equation (I) and solving

t=mv0vdv'F0ev'/v=mF0v0vev'/vdv'=mF0[ev'/v1v]v0v=mF0[ev'/v1v]v0v=mvF0[ev'/v]v0v

t=mvF0[ev/vev0/v]F0tmv=ev/vev0/vev/v=F0tmv+ev0/v

Taking log on both sides

 vv=ln(F0tmv+ev0/v)v(t)=vln(F0tmv+ev0/v)

Conclusion:

Thus, the velocity of the mass with respect to time is v(t)=vln(F0tmv+ev0/v).

(b)

To determine

The time that takes to for the mass to come instantaneously to rest.

(b)

Expert Solution
Check Mark

Answer to Problem 2.14P

The time that takes to for the mass to come instantaneously to rest is t*=mvF0(1ev0/v).

Explanation of Solution

From part (a), the velocity of the mass with respect to time is v(t)=vln(F0tmv+ev0/v). When the mass comes to rest the velocity is zero, v(t)=0.

Therefore,

vln(F0t*mv+ev0/v)=0ln(F0t*mv+ev0/v)=0F0t*mv+ev0/v=e0F0t*mv+ev0/v=1

F0t*mv=1ev0/vt*=mvF0(1ev0/v)

Conclusion:

Thus, the time that takes to for the mass to come instantaneously to rest is t*=mvF0(1ev0/v).

(c)

To determine

The distance the mass travels before coming instantaneously to rest.

(c)

Expert Solution
Check Mark

Answer to Problem 2.14P

The distance the mass travels before coming instantaneously to rest is x(t*)=mv2F0[1ev0v(v0v+1)].

Explanation of Solution

From part (a), the velocity of the mass with respect to time is v(t)=vln(F0tmv+ev0/v). Since velocity is the rate of change of displacement by time, v(t)=ddtx(t).

Therefore, integrating the velocity will gives displacement of the mass.

x(t)=0tv(t')dt'=0tvln(F0t'mv+ev0/v)        (I)

For integration, consider z=F0t'mv+ev0/v. Thus, dz=F0mvdt'  and dt'=mvF0dz.

Solving the integration part in the above equation,

vln(F0t'mv+ev0/v)=lnzmvF0dz=mvF0lnzdz        (II)

Let u=lnz and dv=dz. Using the formula, udv=uvvdu. Here, du=1zdz and v=z.

The integration part becomes,

lnzdz=(lnz)zz(1z)dz=zlnzz

Substitute the above equation in equation (II)

vln(F0t'mv+ev0/v)=mvF0(zlnzz)=mvF0z(lnz1)

Substitute the value of z in the above equation

vln(F0t'mv+ev0/v)=mvF0(F0t'mv+ev0/v)[ln(F0t'mv+ev0/v)1]

Substitute the above integral in equation (I) and solving

x(t)=0tv(t')dt'=vmvF0[(F0t'mv+ev0/v)[ln(F0t'mv+ev0/v)1]]0t=mv2F0[(F0tmv+ev0/v)[ln(F0tmv+ev0/v)1](ev0/v)[ln(ev0/v)1]]=mv2F0[(F0tmv+ev0/v)ln(F0tmv+ev0/v)F0tmvev0/v(ev0/v)ln(ev0/v)+ev0/v]

=mv2F0[(F0tmv+ev0/v)ln(F0tmv+ev0/v)F0tmv(v0vev0v)]=mv2F0(F0tmv+ev0/v)ln(F0tmv+ev0/v)+(mv2F0)F0tmv+(mv2F0)(v0vev0v)=mv2F0(F0tmv+ev0/v)ln(F0tmv+ev0/v)+vt+(mv2F0)(v0vev0v)x(t)=mv2F0[(F0tmv+ev0/v)ln(F0tmv+ev0/v)+v0vev0v]+vt

From part (b), the time that takes to for the mass to come instantaneously to rest is t*=mvF0(1ev0/v).

Therefore,

x(t*)=mv2F0[(F0mv(mvF0(1ev0/v))+ev0/v)ln(F0mv(mvF0(1ev0v))+ev0/v)+v0vev0v]+v(mvF0(1ev0/v))=mv2F0[(1ev0/v+ev0/v)ln(1ev0/v+ev0/v)+v0vev0v1+ev0v]=mv2F0[(1)ln(1)+v0vev0v1+ev0v]=mv2F0[1v0vev0vev0v]x(t*)=mv2F0[1ev0v(v0v+1)]

Conclusion:

Thus, the distance the mass travels before coming instantaneously to rest is x(t*)=mv2F0[1ev0v(v0v+1)].

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A one-dimensional harmonic oscillator of mass m and angular frequency w is in a heat bath of temperature T. What is the root mean square of the displacement of the oscillator? (In the expressions below k is the Boltzmann constant.) Select one: ○ (KT/mw²)1/2 ○ (KT/mw²)-1/2 ○ kT/w O (KT/mw²) 1/2In(2)
Two polarizers are placed on top of each other so that their transmission axes coincide. If unpolarized light falls on the system, the transmitted intensity is lo. What is the transmitted intensity if one of the polarizers is rotated by 30 degrees? Select one: ○ 10/4 ○ 0.866 lo ○ 310/4 01/2 10/2
Before attempting this problem, review Conceptual Example 7. The intensity of the light that reaches the photocell in the drawing is 160 W/m², when 0 = 18°. What would be the intensity reaching the photocell if the analyzer were removed from the setup, everything else remaining the same? Light Photocell Polarizer Insert Analyzer
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON