A spring is mounted horizontally on an air track, with the left end held station

Question

A spring is mounted horizontally on an air track, with the left end held stationary. When a force of 6.0 N is applied, pulling the spring to the right (in the positive x direction), the elongation of the spring is 0.030 m from its equilibrium position. Suppose a 0.50 kg mass is now attached to the right end of the spring, is pulled to a distance of 0.040 m and released. The mass is now oscillating with simple harmonic motion. Find (a) the force constant of the spring (b) the maximum velocity attained by the vibrating mass (c) the maximum acceleration of the vibrating mass (d) the velocity and acceleration when the object has moved halfway to the center from its initial position (i.e. half way to equilibrium position) and (e) the kinetic energy, the potential energy, and the total energy in the halfway position.

Explanation / Answer

a) F = kx, (6.0N) = k(.03 m)   therefore, k= 20 N/m

b) 1/2kx2 = 1/2mv2 evaluate the variables

so, 0.5(20N/m)(0.04m)2 = 0.5(0.5 kg)(v2)

v = 0.25 m/s

c) F = ma = kx

(0.5 kg)(a) = (20 N/m)(0.04m)

a = 1.6 m/sec2

d) The velocity and acceleration would be half of the maximum that was calculated, so v = 0.125 m/s and

a = 0.8 m/sec2

e) KE = 1/2mv2 = 0.5(0.5)(0.125)2 = 0.0039 J

PE = 0.0039 J b/c of the conservation of energy laws

TE = EP + EK since it is the half way to equllibrium, so EP = EK so TE = 0.008 J

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