An unusual spring has a restoring force of magnitude f = (2.00 n/m)x + (1.00 n/m2)x2, where x is the stretch of the spring from its equilibrium length. a 3.00-kg object is attached to this spring and released from rest after stretching the spring 1.50 m. if the object slides over a frictionless horizontal surface, how fast is it moving when the spring returns to its equilibrium length?

Answer

given,

mass = 3 kg

x = 1.50 m

Restoring Force

F = (2.00 N/m)x + (1.00 N/m2)x²

work done =

   =

   =

now work done at x = 1.5

   =

   =

  W= 3.375 J

we know

v = 1.5 m/s

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

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