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 = 38.8 J and a maximum displacement from equilibrium of 0.285 m.

(a) What is the spring constant? 955 N/m Correct:

(b) What is the kinetic energy of the system at the equilibrium point? 38.8 J Correct:

(c) If the maximum speed of the block is 3.45 m/s, what is its mass? 6.52 kg Correct:

(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=0.160m. ________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 = 0.285 m on a rough surface so that it loses 11.2 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) Potential energy of the spring = (1/2)KX2
X = 0.285 m
(1/2)K(0.285)2 = 38.8
K = 955 N/m
(b) Kinetic energy = 38.8 J
(c) Kinetic energy = (1/2)mV2 = (1/2)m*3.452 = 38.8
m = 6.52 kg
(d) When the displacement x = 0.16 m
potetntial energy of the spring = (1/2)Kx2 = 0.5*955*0.162 = 12.224 J
Total mechanical energy = 38.8 J
therefore , kinetic energy = 38.8 - 12.224 = 25.776 J
(1/2)mv2 = 25.776
v = 2.812 m/s
(e) Kinetic energy = 25.776 J
(f) Potential energy of the spring = 12.224 J

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