Aspring is stretched 10 cm by a force of 3 n. a mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 n when the velocity of the mass is 5 m/s. if the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10 cm/s, determine its position u at any time t. find the quasi-frequency î¼ and the ratio of î¼ to the natural frequency of the corresponding undamped motion.

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i just did this and got a 100.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j