Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 6, Problem 45CP

A 9.00-kg object starting from rest falls through a viscous medium and experiences a resistive force given by Equation 6.2. The object reaches one half its terminal speed in 5.54 s. (a) Determine the terminal speed. (b) At what time is the speed of the object three-fourths the terminal speed? (c) How far has the object traveled in the first 5.54 s of motion?

(a)

Expert Solution
Check Mark
To determine

The terminal speed of the object.

Answer to Problem 45CP

The terminal speed of the object is 78.34m/s .

Explanation of Solution

Given information:

The mass of the object is 9.00kg , the time to reach its terminal speed is 5.54sec .

The terminal speed of the object is given as,

vT=mgb (I)

  • m is the mass of the object.
  • b is a constant.
  • g is the acceleration due to gravity.
  • vT is the terminal speed.

The speed of the object at an instant of time is given as,

v=vT(1ebtm) (II)

  • t is the time take.
  • v is the instantaneous speed.

From the given condition, it is clear that the speed of the object is one half of its terminal speed at t=5.54sec .

Therefore, the velocity of the object is,

v=12vT

Substitute 12vT for v in equation (II).

12vT=vT(1ebtm)12=1ebtmebtm=12

Further solve the above expression.

btm=ln(12)mb=tln(2)

Substitute tln(2) for mb in equation (I).

vT=(tln(2))g

Substitute 5.54sec for t and 9.8m/s2 for g in the above equation.

vT=(5.54secln(2))(9.8m/s2)=78.34m/s

Conclusion:

Therefore, the terminal speed of the object is 78.34m/s .

(b)

Expert Solution
Check Mark
To determine

The time at which the speed of the object is three-fourth of the terminal speed.

Answer to Problem 45CP

The time at which the speed of the object is three-fourth of the terminal speed is 11.07sec .

Explanation of Solution

Given information:

The mass of the object is 9.00kg , the time to reach its terminal speed is 5.54sec .

From the given condition, it is clear that the speed of the object is three-fourth of its terminal speed.

Therefore, the velocity of the object is,

v=34vT

Substitute 34vT for v in equation (II).

34vT=vT(1ebtm)34=1ebtmebtm=14btm=ln(14)

Rearrange the above expression for t .

t=(mb)ln(0.25)

Substitute 5.54secln(2) for mb in above equation.

t=(5.54secln(2))ln(0.25)=11.07sec

Conclusion:

Therefore, the time at which the speed of the object is three-fourth of the terminal speed is 11.07sec .

(c)

Expert Solution
Check Mark
To determine

The distance travelled by the object in first 5.54sec of motion.

Answer to Problem 45CP

The distance travelled by the object in first 5.54sec of motion is 120.9m .

Explanation of Solution

Given information:

The mass of the object is 9.00kg , the time to reach its terminal speed is 5.54sec .

The speed of an object is given as,

v=drdt

  • r is the distance travelled by the object.

Rearrange the above expression for r .

r=0tvdt (III)

The speed of the object is given as,

v=vT(1ebtm)

Substitute vT(1ebtm) for v in equation (III).

r=0tvT(1ebtm)dt=vT[t+mb(ebtm1)]

Substitute 78.34m/s for vT , 5.54secln(2) for mb , and 5.54sec for t in above expression.

r=(78.34m/s)[5.54sec+5.54secln(2)(eln(2)(5.54sec)5.54sec1)]=120.9m

Conclusion:

Therefore, the distance travelled by the object in first 5.54sec of motion is 120.9m .

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Chapter 6 Solutions

Physics for Scientists and Engineers with Modern Physics

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