Question

how do I set up last three equations?

A 0.72 kg mass is attached to a light spring with a force constant of 36.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following. (a) maximum speed of the oscillating mass 0.358 m/s (b) speed of the oscillating mass when the spring is compressed 1.5 cm m/s (c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position m/s (d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value m

Answer 1

yoven m=0.72 kg K= 36.9 Nim @ A=0.oşan maximuon speed Vmax = WA= K n WA=W Ja = (17.)x0.06 Vmax= 0.358 m/s fra (6 Total Energy = Umax = te KAZ Sex 36.9% (0.05]? = 0.0465 At, x= 1.5cm à x 36.9% (0.015)2 = 0.0041 I change 1 / mu?=0.0425 ink. E = 0.046-0 0041=0.0425 V? 11 2 x 0.042 m 2x0.042 0 72 = 0.1162 v=0.340g mis tvo V=0.341on's to

At Equilibiriem velocrity is same as part ie VE 0.341 mis AR @ If the speed is is 1 of the max. value 2 Then. kE ish of the max. value & spring Energy 2/3 of maxianyas aralla is 3 so, U ( 2 ) x 0.046= t K x2 0.069=kXL X²= 0.069 k 0.069 36-9 X=0.043 m X=0.043 n Az
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