A10.00 kg mass is attached to a 250n/m spring and set into vertical oscillation. when the mass is 0.50m above the equilibrium it is moving at speed 2.4 m/s. find the amplitude of the oscillation

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

did yall get answers for this yet