Answer:
sqrt(.9576) = v_f
Explanation:
Assuming no friction or air resistance for this.
We can use use conservation of energy for this. Since height isn't changing gravitational potential energy isn't changing, but spring potential energy is.
KE_i + SE_i = FE_f + SE_f
KE is kinetic energy and spring potential energy is SE Initially the mass is not moving so KE_i =0 then SE_i is calculated by .5kx where k is that spring constant and x is the distance from the equilibrium point, so let's plug everything in.
KE_i + SE_i = KE_f + SE_f
.5m(v_i)^2 + .5k(x_i)^2 = .5m(v_f)^2 + .5k(x_f)^2
.5*.3(0)^2 + .5*26.6*(.12)^2 = .5*.3(v_f)^2 + .5*26.6*(.06)^2
.5*26.6*(.12)^2 = .5*.3(v_f)^2 + .5*26.6*(.06)^2
.19152 = .5*.3(v_f)^2 + .04788
.14364 = .5*.3(v_f)^2
.9576 = (v_f)^2
sqrt(.9576) = v_f
Let me know if you couldn't follow the algebra.
sqrt(.9576) = v_f is the speed of the mass when it is 0.0600 m from equilibrium
KE_i + SE_i = FE_f + SE_f
KE_i + SE_i = KE_f + SE_f
.5m(v_i)^2 + .5k(x_i)^2 = .5m(v_f)^2 + .5k(x_f)^2
.5*.3(0)^2 + .5*26.6*(.12)^2 = .5*.3(v_f)^2 + .5*26.6*(.06)^2
.5*26.6*(.12)^2 = .5*.3(v_f)^2 + .5*26.6*(.06)^2
.19152 = .5*.3(v_f)^2 + .04788
.14364 = .5*.3(v_f)^2
.9576 = (v_f)^2
sqrt(.9576) = v_f
KE is kinetic energy and spring potential energy is SE Initially the mass is not moving so KE_i =0 then SE_i is calculated by .5kx where k is the spring constant and x is the distance from the equilibrium point, so let's plug everything in.
Learn more about equilibrium here: https://brainly.com/question/14297698
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