A 1 kg object is attached to a horizontal spring. It is stretched by 0.1 m and released from rest. It moves without friction. The next time the speed is zero is 0.5 s later. What is the maximum speed of the object

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