A 7.5 kg block is attached to a horizontally mounted spring with a spring constant of 750 J/m2 . The block is released from rest with the spring extended by 15 cm from its normal length. The block then oscillates back and forth along a horizontal and frictionless surface. How long will it take this system to complete one full oscillation?

It will take the system 0.63sec to complete one full oscillation

Explanation:

The expression for period of a body in simple harmonic motion is

T=2*pi√m/k

Where T Is the period

m is the mass of the body

K is the spring constant

Given data

m = 7.5kg

e=15cm

k=750j/m2

Substituting our data into the expression

T=2*3.142 √ 7.5/750

T=6.284*0.1

T=0.63sec

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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