A particle P of mass 5 kg is released from rest in the position shown and slides down the track with negligible friction until it hits a spring at x = 2m. Knowing that the spring is linear and has a constant k = 400 N / m. Determine the velocity of the particle when the spring has been compressed a distance of 0.8 m.

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

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