Posted by Anonymous on Monday, November 26, 2012 at 4:15pm.
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(A²-x²)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=energyat4.2-energyat4.0
= 1/2 k (4.2^2 -4^2)
then knowing KE, solve for velocity
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