Answer:
Explanation:
The frictionless surface implies that the speed of the spring is at a max. When the speed of the spring is at its max, the potential energy in the spring is 0. Use the equation for the Total Energy in a Spring/Mass System:
KE + PE = [tex]\frac{1}{2}kA^2[/tex] where KE is the Kinetic Energy available to the spring, PE is the potential energy available to the spring, and the sum of those is equal to one-half times the spring constant, k, times the amplitude of the spring's movement away from the equilibrium position. Sometimes this amplitude is the same as the displacement of the spring. This can be tricky. But since we are only given one value for the distance, we are going to use it as an amplitude. Keeping in mind that the PE is 0 when KE is at its max, then the equation becomes
KE + 0 = [tex]\frac{1}{2}kA^2[/tex] or to put it simpler terms:
KE = [tex]\frac{1}{2}kA^2[/tex] We need to find the value for KE before we can fully solve the problem we are being tasked with.
Filling in using the info given:
[tex]KE=\frac{1}{2}(2.0)(.06)^2[/tex] Notice I added another place of significance to the 2 because 1 simply isn't enough and the physics teacher in me can't handle that. Simplifying a bit:
[tex]KE=(.06)^2[/tex] because the k = 2 cancels out the 2 in the denominator of the 1/2. So
KE = 3.6 × [tex]10^{-3[/tex]
Now plug that in for KE and solve for v:
KE = [tex]\frac{1}{2}mv^2[/tex]:
[tex]3.6*10^{-3}=\frac{1}{2}(5.0)v^2[/tex] and
[tex]v=\sqrt{\frac{2(3.6*10^{-3})}{5.0} }[/tex] gives us a velocity of
v= [tex]3.8*10^{-2[/tex]
