A horizontal block-spring system with the block on a frictionless surface has to

Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 51.7 J and a maximum displacement from equilibrium of 0.262 m.
(a) What is the spring constant?
N/m

(b) What is the kinetic energy of the system at the equilibrium point?
J

(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.
J

(f) Find the potential energy stored in the spring when x = 0.160 m.
J

(g) Suppose the same system is released from rest at x = 0.262 m on a rough surface so that it loses 11.4 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?
m

Explanation / Answer

a) 1/2kx2 = 51.7; x = 0.262 => k = 1506.32 N/m b) KE = All of Spring energy = 51.7 J c) 1/2mv2 = 51.7; v = 3.45m/s => m = 8.69 kg d) When displacement is 0.16, Spring energy = 19.28 J KE = 31.62 J Velocity= 2.7 m/s e) KE = 31.62 J f) PE in spring = 19.28 J g) Energy left = 51.7-11.4 = 40.3J x = 0.231 m

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