Secure | https://sessionmasteringp temView?assignmentProblemlD-822014008 offset

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Secure | https://sessionmasteringp temView?assignmentProblemlD-822014008 offset . Assignment 29 Problem 15.18- Enhanced - with Feedback previous 3 of 4 | next Problem 15.18 Enhanced - with Feedback Part A The amplitude of the subsequent oscillations? A 1.15 kg block is attached to a spring with spring constant 15.5 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 440 cm/s What are Express your answer with the appropriate units. You may want to review/Prages400-401) 64.68 cm Subrnit My Answers Give Up Incorrect, Try Again; 4 attempts remaining Part B The block's speed at the point where r 0.100 A? Express your answer with the appropriate units. Type here to search

Explanation / Answer

Part A:

Here the initial kinetic energy of the block shall be equal to the energy stored in the spring as potential energy.

now -  
Ekin = 1/2 mv^2 = 1/2*1.15*0.44^2

and this is equal to the spring energy 1/2 kA^2

1/2*1.15*0.44^2 = 1/2*15.5*A^2
1.15*0.44^2 = 15.5*A^2
A = 0.12 m = 12.0 cm

Hence the amplitude of oscillation = 12.0 cm

Part B -

The equation of the motion is
y = 0.12sin((k/m)*t)
and v = 1. derivative
v = 0.12*(k/m)cos((k/m)t)

find t when y = 0.10 A = 0.012 m
0.012 = 0.12*sin((15.5/1.15)t)
0.10 = sin(3.67t)
3.67t = 0.10
=> t = 0.10 / 3.67 = 0.02725 s

now put this into above -  
y = 0.12sin((15.5/1.15)*0.02725)  
v = 0.012 m/s = 1.20 cm/s

Please note that I have calculated the variables with a high level of concentration. However, I request you to please check once more from your end.

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