Answer:
v = 8.46 m/s
Explanation:
Given that,
Mass of the object, m = 14.6 g
Spring constant of the spring, k = 15.7 N/m
It is released from rest with an amplitude of 29.8 cm.
We need to find the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless.
We can use the conservation of energy in this case. Let v be the speed of the mass. So,
[tex]EPE_i=\dfrac{1}{2}mv^2+EPE_f\\\\\dfrac{1}{2}kx_i^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx_f^2[/tex]
Substitue all the values in the above expression.
[tex]\dfrac{1}{2}\times 15.7\times (0.298)^2=\dfrac{1}{2}\times 0.0146\times v^2+\dfrac{1}{2}\times 15.7\times (\dfrac{0.298}{2})^2\\\\v=8.46\ m/s[/tex]
So, the speed of the mass is equal to 8.46 m/s.
