Answer:
K=0.4368J
Explanation:
The mechanical energy will be the sum of the elastic energy and the kinetic energy of the block: [tex]E=U+K[/tex]. Their formulas are [tex]U=\frac{k\Delta x^2}{2}[/tex] and [tex]K=\frac{mv^2}{2}[/tex] for any point in the movement (although E must remain constant since the movement is frictionless), where k is the spring constant, [tex]\Delta x[/tex] the displacement, m the mass of the block and v its velocity.
When the block is released from rest, at that point its mechanical energy is equal to the elastic energy, which is at its maximum ([tex]E=U_{max}[/tex]) since the displacement is at its maximum ([tex]\Delta x_{max}=0.08m[/tex]).
Since mechanical energy is conserved, the mechanical energy at that point must be equal also to the mechanical energy at the point asked, which we will call P (when the displacement is [tex]x_P=0.024m[/tex]), so we will have:
[tex]E=U_{max}=U_P+K_P[/tex]
So the kinetic energy at point P is
[tex]K_P=U_{max}-U_P=\frac{k \Delta x_{max}^2}{2}-\frac{k \Delta x_P^2}{2}=\frac{k (\Delta x_{max}^2-\Delta x_P^2)}{2}[/tex]
And substituting values we have:
[tex]K_P=\frac{(150N/m) ((0.08m)^2-(0.024m)^2)}{2}=0.4368J[/tex]
