## Question

###### 0.190 m and the block moves a 3.50 m/s as it passes through A 0.325 kg...

0.190 m and the block moves a 3.50 m/s as it passes through A 0.325 kg block attached to a light spring osciates on a frictionless, horizontal table. The station amplitude is equilibrium at- HINT (in N/m). (a) Find the spring constant, N/m (D) Calculate the total energy ( ) of the block spring system, (c) Find the block's speed (in m/s) when -

## Answers

1 For the sum of a mass-sphing eystem displacement is inz Aco (wt + $5 where Ar amplitude - er engulas frequeny velocity ve du ve - Asin (ast 16). Ve- Aw sin (wt4).we have sin a + cord liels va ofw ( KY . Given : 3,5 mis when VZDA (1-0) " 3.5 0.19 w. 2 19.42 hells ay we have worse 2 where kis spring constant kz 0.325 (18.42)2 . kz 110.27 Nlmb otg goevg a SUM Ez I KnewAd 10-27X6.1972 E - 1.99014 J C) Cinen K A speed, vo Ano (- (Al) na V = 0.19x18-4a (1 – t)s V: 3.499 (2) va 3.03 m/s

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