A 1.20-kg cart connected to a light spring forwhich the force constant is 20.0 N/m oscillates onamaxa frictionless, horizontal air track.(A) Calculate the maximum speed of the cart ifthe amplitude of the motion is 3.00 cm.(B) What is the velocity of the cart when theposition is 2.00 cm?amax(C) Compute the kinetic and potential energies ofthe system when the position of the cart is2.00 cm.maxSOLVE IT(A) Calculate the maximum speed of the cart ifamaxthe amplitude of the motion is 3.00 cm.Conceptualize The system oscillates in exactlythe same way as the block in the figure.Categorize The cart is modeled as a particle insimple harmonic motion.-A-ASeveral instants in the simple harmonic motion fora block-spring system.11 Analyze Use the equation to express the1 KA? = 1mv?.maxtotal energy of the oscillator system and2equate it to the kinetic energy of thesystem when the cart is at x = 0:Solve for the maximum speed:kA =V.maxm/sm(B) What is the velocity of the cart when the position is 2.00 cm?Use the equation to evaluate the velocity:k(A² - x3)V = ±m20.0 N/m| (0.0300 m)2 - (0.0200 m)리1.20 kg= ±m/sThe positive and negative signs indicate that the cart could be moving to either the right or the left atthis instant.(C) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm.Use the result of part (B) to evaluate thekinetic energy at x =글까2-흑(1.20 kg)v2-K ==0.0200 m:Evaluate the elastic potential energy atU =kx² = (20.0 N/m)(0.0200 m)2 :X = 0.0200 m:2Notice that the sum of the kinetic and potential energies in part (C) is equal to the total energy found inpart (A). That must be true for any position of the cart.MASTER ITHINTS:GETTING STARTED I I'M STUCK!When the mass is at the center and at rest, we give it a kick and increase its velocity to 0.305 m/s.(a) What is the new amplitude (in cm) of the oscillation now?A =cm(b) When the mass is at -2.7 cm, what is its speed (in m/s)?V =m/s

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Description: A 1.20-kg cart connected to a light spring forwhich the force constant is 20.0 N/m oscillates onamaxa frictionless, horizontal air track.(A) Calculate the maximum speed of the cart ifthe amplitude of the motion is 3.00 cm.(B) What is the velocity of

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(C) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm.

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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Topic Video

Question

A 1.20-kg cart connected to a light spring for
which the force constant is 20.0 N/m oscillates on
amax
a frictionless, horizontal air track.
(A) Calculate the maximum speed of the cart if
the amplitude of the motion is 3.00 cm.
(B) What is the velocity of the cart when the
position is 2.00 cm?
amax
(C) Compute the kinetic and potential energies of
the system when the position of the cart is
2.00 cm.
max
SOLVE IT
(A) Calculate the maximum speed of the cart if
amax
the amplitude of the motion is 3.00 cm.
Conceptualize The system oscillates in exactly
the same way as the block in the figure.
Categorize The cart is modeled as a particle in
simple harmonic motion.
-A
-A
Several instants in the simple harmonic motion for
a block-spring system.
11

Analyze Use the equation to express the
1 KA? = 1mv?.
max
total energy of the oscillator system and
2
equate it to the kinetic energy of the
system when the cart is at x = 0:
Solve for the maximum speed:
k
A =
V.
max
m/s
m
(B) What is the velocity of the cart when the position is 2.00 cm?
Use the equation to evaluate the velocity:
k
(A² - x3)
V = ±
m
20.0 N/m
| (0.0300 m)2 - (0.0200 m)리
1.20 kg
= ±
m/s
The positive and negative signs indicate that the cart could be moving to either the right or the left at
this instant.
(C) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm.
Use the result of part (B) to evaluate the
kinetic energy at x =
글까2-흑(1.20 kg)v2-
K =
=
0.0200 m:
Evaluate the elastic potential energy at
U =
kx² = (20.0 N/m)(0.0200 m)2 :
X = 0.0200 m:
2
Notice that the sum of the kinetic and potential energies in part (C) is equal to the total energy found in
part (A). That must be true for any position of the cart.
MASTER IT
HINTS:
GETTING STARTED I I'M STUCK!
When the mass is at the center and at rest, we give it a kick and increase its velocity to 0.305 m/s.
(a) What is the new amplitude (in cm) of the oscillation now?
A =
cm
(b) When the mass is at -2.7 cm, what is its speed (in m/s)?
V =
m/s

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