A 0.35-kg object connected to a light spring with a force constant of 23.2 N/m o

Question

A 0.35-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest.
(a) Determine the maximum speed of the object.
.
Your response is off by a multiple of ten. cm/s

(b) Determine the speed of the object when the spring is compressed 1.5 cm.
.
Your response is off by a multiple of ten. cm/s

(c) Determine the speed of the object when the spring is stretched 1.5 cm.
.
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm/s

(d) For what value of x does the speed equal one-half the maximum speed?

Explanation / Answer

m = 0.35 kg

k = 23.2 N/m

maximum compression = 4cm = 0.04 m

potential energy stored in the spring is converted to kinetic energy

a) 0.5kx2 = 0.5 mv2

=> v = x(k/m) = 0.04*(23.2/0.35) = 0.326 m/s = 3.26 cm/s

b)0.5 * k * 0.042 = 0.5* k * 0.0152 + 0.5 * 0.35 * v2

=> 0.35 * v2= k * (0.042-0.0152)

=> v = 0.301 m/s = 3.01 cm/s

c) the answer will be same for stretching 1.5 cm

d)0.5 * 23.2 * 0.042 = 0.5 * 23.2 * x2 +0.5 * 0.35 * (0.326/2)2

=> 0.75 * 23.2 * 0.042 = 23.2 * x2 since (0.5 * 0.35 * (0.326/2)2 = 0.75*0.5 * 23.2*0.042)

=> x = 0.0346 m = 0.346 cm

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