physics
posted by Anonymous .
A cart of mass 223 g is placed on a frictionless horizontal air track. A spring having a spring constant of 10.20 N/ m is attached between the cart and the left end of the track. When in equilibrium, the cart is located 11.0 cm from the left end of the track. If the cart is displaced 4.20 cm from its equilibrium position, find
(c) its speed when it is 14.0 cm from the left end of the track.

When the cart is 14 cm from the left end of the track, it has a displacement of
x=14 11 cm =3 cm= 0.03 m from the equilibrium position.
The speed of the cart at this distance from equilibrium is
v=sqrt{k(A²x²)m}=… 
ok, it released spring energy 0.2cm from the starting position to the mark.
energyreleased=INT force*dx from 4.2 to 4.0 cm if you are a calculus person.
for non calculcus persons the KE it has is the difference of spring PE
KE=energyat4.2energyat4.0
= 1/2 k (4.2^2 4^2)
then knowing KE, solve for velocity