To solve this problem we will apply the concepts related to the conservation of kinetic energy and elastic potential energy. Thus we will have that the kinetic energy is
[tex]KE = \frac{1}{2} mv^2[/tex]
And the potential energy is
[tex]PE = \frac{1}{2} kx^2[/tex]
Here,
m = mass
v = Velocity
x = Displacement
k = Spring constant
There is equilibrium, then,
KE = PE
[tex]\frac{1}{2} mv^2 = \frac{1}{2} kx^2[/tex]
Our values are given as,
[tex]x=0.56m\\k=230N/m\\m=15kg[/tex]
Replacing we have that
[tex]\frac{1}{2} (15)v^2 = \frac{1}{2} (230)(0.56)^2[/tex]
[tex]v = 2.19m/s[/tex]
Therefore the speed of the cart is 2.19m/s
