Question

please fix these to get the -4 points

just the question need to be fix. thanks

Sally inward my? ng eroro Solve following problems. You must show your work to earn credits. 1. A 3.00 kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 900 N/m. The spring initially is neither stretched nor compressed. The fish is released from rest. (a) What is its speed after it has descend 0.0500m from its initial position? (b) What is the maximum speed of the fish as it descends? 1 Going back to the formula forv, we Ponential energy = mgs see that: Kinetic energy Imre V V ( 2 (mgs - 1 x Vis?) /m). P erergy IKS 2 At the rest point is the polontial energy relative V= v(29s - ks 2 lm) ? to the position 0.05m lower amgs = x 9.81 +0.055. V=V (2 x 9.8 ls - 300s ²) Kinetic energy=0 Spring energy o "Now we look for the maximum I of this function: After releasing and 0.05 m lower is the potential the derivative is energy-o Kinetic energy - x 3 x v2J Tv'- (981-3005) /u (s (1962-2005) spring energy - 1 x 900 x d.os? so 981-800s x S- mg and Eput tkin & Espre 3:321 (3.2cm) 3 x 9,816 dosyx3x vett x 900 x 0.052/ 10000 - v=0.406069 mis (speed after 5cm) (max) = 327v3/1000 = 0.566381 mis (max speed) Het perfectly Glash caftsin

Answer 1

(9) ALE 00500 m. uitk, I U2+k2. (3) ng (0) + I k (0)² + 2 (0)2 mg (-4L) + ₂ K (A6) ² + 2 mu? ngal - ₂ K OL² = 1 my 2 2mg AL-KAL VI ((2x3x9.840.0500) - 900 (0.0500) m V2 = 0.480 mis Express V2 as a function of oL V22 Jagal-kol 2 (2 g OL- 1 62² / ² find vala) = 0 V₂ (AL) = 1 (29 0 1 - K Ligh (29- 2 KAL) = 0 V₂ (AL) - 29- 2K OL 29- 26 AL to 2 oLam AL = 3x9..8 = 0.0327 ml V₂ (0.0327) = 2 x 9.87 0.0327 - 900 (0.0327) ² 3.00 V2 max z 0.566 m/s.