Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
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Chapter 15, Problem 15.28P

A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. Find (a) the force constant of the spring, (b) the frequency of the oscillations, and (c) the maximum speed of the object, (d) Where does this maximum speed occur? (e) Find the maximum acceleration of the object. (f) Where does the maximum acceleration occur? (g) Find the total energy of the oscillating system. Find (h) the speed and (i) the acceleration of the object when its position is equal to one-third the maximum value.

(a)

Expert Solution
Check Mark
To determine

The force constant of the spring.

Answer to Problem 15.28P

The force constant of the spring is 100N/m .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

Write the expression for force constant.

k=|F||A|

Here,

F is the force on the spring.

A is the amplitude.

Substitute 0.200m for A and 20.0N for F in above equation.

k=|20.0N||0.200m|=100N/m

Conclusion:

Therefore, the force constant of the spring is 100N/m .

(b)

Expert Solution
Check Mark
To determine

The frequency of the oscillations.

Answer to Problem 15.28P

The frequency of the oscillations is 1.125Hz .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

Write the expression for the force constant of the spring.

k=mω2ω=km

Here,

k is the force constant.

m is the mass of the spring.

ω is the angular velocity.

Substitute 100N/m for k and 2.00kg for m in above equation.

ω=100N/m2.00kg=50rad/s

Write the expression for the frequency of the oscillations.

f=ω2π

Here,

f is the frequency of the oscillations.

Substitute 50rad/s for ω in above equation.

f=50rad/s2π=1.125Hz

Conclusion:

Therefore, the frequency of the oscillations is 1.125Hz .

(c)

Expert Solution
Check Mark
To determine

The maximum speed of the object.

Answer to Problem 15.28P

The maximum speed of the object is 1.41m/s .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

From part (b) the angular velocity is 50rad/s .

Write the expression for maximum speed.

vmax=ωA

Here,

vmax is the maximum speed.

Substitute 50rad/s for ω and 0.200m for A in above equation.

vmax=50rad/s×0.200m=1.41m/s

Conclusion:

Therefore, the maximum speed of the object is 1.41m/s .

(d)

Expert Solution
Check Mark
To determine

The position of the object where the maximum speed occurs.

Answer to Problem 15.28P

The position of the object where the maximum speed occurs at x=0 .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

The maximum speed of the object occurs when the object passes through its equilibrium position.

The equilibrium position of the object is,

x=0

Conclusion:

Therefore, the position of the object where the maximum speed occurs at x=0 .

(e)

Expert Solution
Check Mark
To determine

The maximum acceleration of the object.

Answer to Problem 15.28P

The maximum acceleration of the object is 10m/s2 .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

Write the expression for the maximum acceleration.

amax=ω2A

Here,

amax is the maximum acceleration.

Substitute 50rad/s for ω and 0.200m for A in above equation.

amax=(50rad/s)2(0.200m)=10m/s2

Conclusion:

Therefore, the maximum acceleration of the object is 10m/s2 .

(f)

Expert Solution
Check Mark
To determine

The position of the object where the maximum acceleration occurs.

Answer to Problem 15.28P

The position of the object where the maximum acceleration occurs at x=±0.200m .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

The maximum acceleration of the object occurs when the object reverses its direction of motion.

The object reverses its direction of motion where its distance is the maximum from the equilibrium position.

The maximum distance from equilibrium position of the object is,

x=±0.200m

Conclusion:

Therefore, the position of the object where the maximum acceleration occurs at x=±0.200m .

(g)

Expert Solution
Check Mark
To determine

The total energy of the system.

Answer to Problem 15.28P

The total energy of the system is 2J .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

From part (a) the force constant of the spring is 100N/m .

Write the expression for the total energy.

E=kA22

Here,

E is the total energy of the object.

A is the amplitude of the motion.

Substitute 100N/m for k and 0.200m for A in above equation.

E=(100N/m)(0.200m)22=2J

Conclusion:

Therefore, the total energy of the system is 2J .

(h)

Expert Solution
Check Mark
To determine

The speed of the object when the object is at one-third of the maximum value.

Answer to Problem 15.28P

The speed of the object when the object is at one-third of the maximum value is 1.333m/s .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

The position of the object is one-third of the maximum value.

x=A3

Here,

A is the amplitude of the motion.

x is the position of the object.

Substitute 0.200m for A in above equation.

x=0.200m3=0.0667m

Thus, the value of x is 0.0667m .

From part (a) the angular velocity is 50rad/s .

Write the expression for the velocity of block at any position.

v=ωA2x2

Here,

v is the velocity of the object.

Substitute 50rad/s for ω , 0.0667m for x and 0.200m for A in above equation.

v=(50rad/s)(0.200m)2(0.0667m)2=1.333m/s

Conclusion:

Therefore, the speed of the object when the object is at one-third of the maximum value is 1.333m/s .

(i)

Expert Solution
Check Mark
To determine

The acceleration of the object when the object is at one-third of the maximum value.

Answer to Problem 15.28P

The acceleration of the object when the object is at one-third of the maximum value is 3.333m/s2 .

Explanation of Solution

Given info: The force required to hold the object at rest is 20.0N , the amplitude is 0.200m and the mass of the object is 2.00kg .

From part (h) the position of the object is one-third of the maximum value is 0.0667m .

From part (a) the angular velocity is 50rad/s .

Write the expression for the acceleration of block at any position.

a=ω2x

Here,

a is the acceleration of the object.

Substitute 50rad/s for ω and 0.0667m for x in above equation.

a=(50rad/s)2(0.0667m)=3.333m/s2

Conclusion:

Therefore, the acceleration of the object when the object is at one-third of the maximum value is 3.333m/s2 .

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Chapter 15 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 15 - An object of mass 0.40 kg, hanging from a spring...Ch. 15 - A runaway railroad car, with mass 3.0 105 kg,...Ch. 15 - The position of an object moving with simple...Ch. 15 - If an object of mass m attached to a light spring...Ch. 15 - You stand on the end of a diving board and bounce...Ch. 15 - A mass-spring system moves with simple harmonic...Ch. 15 - A block with mass m = 0.1 kg oscillates with...Ch. 15 - For a simple harmonic oscillator, answer yes or no...Ch. 15 - The top end of a spring is held fixed. A block is...Ch. 15 - Which of the following statements is not true...Ch. 15 - A simple pendulum has a period of 2.5 s. (i) What...Ch. 15 - A simple pendulum is suspended from the ceiling of...Ch. 15 - A particle on a spring moves in simple harmonic...Ch. 15 - You are looking at a small, leafy tree. 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A 65.0-kg bungee jumper steps off a bridge...Ch. 15 - Review. A 0.250-kg block resting on a...Ch. 15 - Prob. 15.32PCh. 15 - While driving behind a car traveling at 3.00 m/s,...Ch. 15 - A seconds pendulum is one that moves through its...Ch. 15 - A simple pendulum makes 120 complete oscillations...Ch. 15 - A particle of mass m slides without friction...Ch. 15 - A physical pendulum in the form of a planar object...Ch. 15 - A physical pendulum in the form of a planar object...Ch. 15 - The angular position of a pendulum is represented...Ch. 15 - Consider the physical pendulum of Figure 15.16....Ch. 15 - Prob. 15.41PCh. 15 - A very light rigid rod of length 0.500 m extends...Ch. 15 - Review. A simple pendulum is 5.00 m long. 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