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!
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
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
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Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5630fc0f-1d39-4405-9289-17dd8428ca98%2Ff1625a31-0fca-48b5-b36f-f248f7566144%2Fs2vxwxn_processed.png&w=3840&q=75)
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
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