Use the worked example above to help you solve this problem. A 0.545 kg object connected to a light spring with a spring constant of 19.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = "max = J m/s (b) What is the velocity of the object when the displacement is 2.00 cm? ± m/s (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J PE, = J EXERCISE HINTS: GETTING STARTED I'M STUCK!

College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter16: Oscillatory Motion And Waves
Section: Chapter Questions
Problem 45PE: Suppose you have a 0.750kg object on a horizontal surface connected to a spring that has a force...
icon
Related questions
icon
Concept explainers
Topic Video
Question
PRACTICE IT
Use the worked example above to help you solve this problem. A 0.545 kg object connected to a light
spring with a spring constant of 19.0 N/m oscillates on a frictionless horizontal surface.
(a) Calculate the total energy of the system and the maximum speed of the object if the
amplitude of the motion is 3.00 cm.
E =
J
m/s
V
max
=
(b) What is the velocity of the object when the displacement is 2.00 cm?
±
m/s
(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.
KE =
J
J
PES=
EXERCISE
For what values of x is the speed of the object 0.14 m/s?
x = ±
cm
HINTS: GETTING STARTED I I'M STUCK!
Transcribed Image Text:PRACTICE IT Use the worked example above to help you solve this problem. A 0.545 kg object connected to a light spring with a spring constant of 19.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. E = J m/s V max = (b) What is the velocity of the object when the displacement is 2.00 cm? ± m/s (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE = J J PES= EXERCISE For what values of x is the speed of the object 0.14 m/s? x = ± cm HINTS: GETTING STARTED I I'M STUCK!
STRATEGY The total energy of the system can be found most easily at the amplitude x = A, where the
kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of
energy then yields the speed at x = 0. For part (b), obtain the velocity by substituting the given value of x
into the time independent velocity equation. Using this result, the kinetic energy asked for in part (c) can
be found by substitution, as can the potential energy.
SOLUTION
(A) Calculate the total energy and maximum speed if the amplitude is 3.00 cm.
E = KE + PEg + PES
= 0 + 0 + ¹2kA² = ½(20.0 N/m)(3.00 x 10-² m)²
= 9.00 × 10-³ J
Substitute x = A = 3.00 cm and
k = 20.0 N/m into the equation for the
total mechanical energy E.
Use conservation of energy with x; = A
and
xf
= 0 to compute the speed of the
object at the origin.
Substitute known values.
(KE + PE + PE§); = (KE + PEg + PEs) f
0 + 0 + 1₂kA² = 1/2mv²_
1/2mv²
Substitute into the equation for spring
potential energy.
Vmax
max
V =
= 9.00 × 10-³ J
(B) Compute the velocity of the object when the displacement is 2.00 cm.
v = ± √k/m(A² – x²)
20.0 N/m
0.500 kg
(
= ±0.141 m/s
±
1.80 x 10-² J
0.500 kg
max
= 1⁄2kx²
+0+0
PES
= 4.00 × 10-³ J
0.190 m/s
(C) Compute the kinetic and potential energies when the displacement is 2.00 cm.
Substitute into the equation for kinetic
energy.
=
[(0.0300 m)² (0.0200 m
-
KE = 1/2mv² = 1/2(0.500 kg)(0.141 m/s)² = 4.97 × 10-³ J
²])
(20.0 N/m) (2.00 x 10-² m)²
½
Transcribed Image Text:STRATEGY The total energy of the system can be found most easily at the amplitude x = A, where the kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of energy then yields the speed at x = 0. For part (b), obtain the velocity by substituting the given value of x into the time independent velocity equation. Using this result, the kinetic energy asked for in part (c) can be found by substitution, as can the potential energy. SOLUTION (A) Calculate the total energy and maximum speed if the amplitude is 3.00 cm. E = KE + PEg + PES = 0 + 0 + ¹2kA² = ½(20.0 N/m)(3.00 x 10-² m)² = 9.00 × 10-³ J Substitute x = A = 3.00 cm and k = 20.0 N/m into the equation for the total mechanical energy E. Use conservation of energy with x; = A and xf = 0 to compute the speed of the object at the origin. Substitute known values. (KE + PE + PE§); = (KE + PEg + PEs) f 0 + 0 + 1₂kA² = 1/2mv²_ 1/2mv² Substitute into the equation for spring potential energy. Vmax max V = = 9.00 × 10-³ J (B) Compute the velocity of the object when the displacement is 2.00 cm. v = ± √k/m(A² – x²) 20.0 N/m 0.500 kg ( = ±0.141 m/s ± 1.80 x 10-² J 0.500 kg max = 1⁄2kx² +0+0 PES = 4.00 × 10-³ J 0.190 m/s (C) Compute the kinetic and potential energies when the displacement is 2.00 cm. Substitute into the equation for kinetic energy. = [(0.0300 m)² (0.0200 m - KE = 1/2mv² = 1/2(0.500 kg)(0.141 m/s)² = 4.97 × 10-³ J ²]) (20.0 N/m) (2.00 x 10-² m)² ½
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps

Blurred answer
Knowledge Booster
Simple Harmonic Motion
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers with Modern …
Physics for Scientists and Engineers with Modern …
Physics
ISBN:
9781337553292
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Glencoe Physics: Principles and Problems, Student…
Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill