MasteringPhysics: Hw4: Chapter 14.1 to 14.5 -Google Chrome Secure https://sessio

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MasteringPhysics: Hw4: Chapter 14.1 to 14.5 -Google Chrome Secure https://session mastering physics.com/myct/itemview?assignmentProblemID 75126773 RUPHYS124Spr2017 HW4: Chapter 14.1 14.5 Exercise 14.26 Exercise 14.26 A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. Correct The amplitude of the motion is 0.290 m and the period is 3.39 s. Part B What is the speed of the block when z 0.160 m Express your answer with the appropriate units. 0.1084 m Submit My Answers Gve Up Incorrect; Try Again; 4 attempts remaining

Explanation / Answer

Using law of conservation of energy

Total energy at the 0.29m = Total energy at the x = 0.16 m

0.5*k*A^2 = (0.5*m*v^2) + (0.5*k*x^2)

since springconstant is k = m*w^2

0.5*m*w^2*A^2 = (0.5*m*v^2) + (0.5*m*w^2*x^2)

m cancels

0.5*w^2*A^2 = (0.5*v^2) + (0.5*w^2*x^2)

Period is T =2*pi/w

w = (2*pi)/T = (2*3.142)/3.39 = 1.85 rad/s

then

0.5*1.85^2*0.29^2 = (0.5*v^2)+(0.5*1.85^2*0.16^2)

V = 0.447 m/sec

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