suhe total energy of a block-spring system is 0.2 J. The block oscillates with an angulal frequency of 20 rad/s and amplitude of 10 cm. Find (a) Find the mass of the block, (b) Find the maximum speed, (c) Find the kinetic energy when the displacement is at 6 cm, (d) Find the positions when kinetic energy equals potential energy. 0.0G m
suhe total energy of a block-spring system is 0.2 J. The block oscillates with an angulal frequency of 20 rad/s and amplitude of 10 cm. Find (a) Find the mass of the block, (b) Find the maximum speed, (c) Find the kinetic energy when the displacement is at 6 cm, (d) Find the positions when kinetic energy equals potential energy. 0.0G m
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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Question

Transcribed Image Text:SThe total energy of a block-spring system is 0.2 J. The block oscillates with an
angulal frequency of 20 rad/s and amplitude of 10 cm. Find
(a) Find the mass of the block,
(b) Find the maximum speed,
(c) Find the kinetic energy when the displacement is at 6 cm,
(d) Find the positions when kinetic energy equals potential energy.
0.0G M
Expert Solution
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Step 1
Given E=0.2 J
A= 10 cm = 0.1 m
= 20 rad/s
a) = (k/m)0.5
E= kinetic energy + potential energy
when the displacement of the body is maximum kinetic enrgy is zero
Therfore potential enrgy = 0.2 J when x=A
P.E= 0.5kx2
0.2= 0.5*k*0.1*0.1
k= 0.2/0.5*0.01
k =40 N/m
m= k/2
= 40/20*20
m= 0.1 Kg
b) For max speed x=0
E= 0+K.E
K.E= 0.5mv2
0.2=0.5*0.1*v2
v2= 4
V(max)= 2m/sec
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