Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 53.7 ) and a maximum displacement from equilibrium of 0.200 m. (a) What is the spring constant? N/m (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? | kg (d) What is the speed of the block when its displacement is 0.160 m? m/s (e) Find the kinetic energy of the block at x = 0.160 m. (1) Find the potential energy stored in the spring when x = 0.160 m. (9) Suppose the same system is released from rest at x = 0.200 m on a rough surface so that it loses 13.6) by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant? m

Answer 1

24 (53.7 J) =ļk(0.200m)" k=/2685 N/m At the equ. pt., KE = Total enegy 53.7 J LE v2 2 I needed) E- ļ ka? 19.02 key] energy 2 =?v=12.07 mis 2x83.7 3.452 E= 1 mu?t I kn? { mv²= E - Į kn?. ₂ (9.02) v2 = 53.7 Ź (2685) (0 le 232 mis) KE = mu? = 1/2 x 9.02% 2,022 =/ 19.3] & PE= E- KE=53.7- 19.3] 34.37 J 8 51.7 - 13.6 = you I J thoning pt, bez v=o Etotal REG = 40.15 į ku2a yo. 1 on Ź (2685) = x=10.123 m) Cuse E-S37 J total the the twring pt. KE=0 at the . Yool

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