A small mailbag is released from a helicopter that is descending steadily at 1.95 m/s. (a) After 3.00 s, what is the speed of the mailbag?
(b) How far is it below the helicopter?

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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