Question

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force constant 5.5 N/m . You pull on the glider, stretching the spring 0.54 m from the equilibrium point, and then release it with no initial velocity. The glider begins to move back toward its equilibrium position (x=0).What is its x-velocity when x = 0.080 m?

Answer 1

`v=2.556m/s`

Step-by-step explanation:

From the conservation of mechanical energy

`K_(E1)+U_1=K_(E2)+U_2`

`(1)/(2)m*v_1^2+(1)/(2)*K*x_1^2=(1)/(2)m*v_2^2+(1)/(2)*K*x_2^2`

`x_2=0.08m`

`v_1=0 m/s`

Solve to velocity v2

`m*v_2^2=k*x_1^2-k*x_2^2`

`v^2=(k)/(m)*(x_1^2-x_2^2)`

`v^2=(5.5N/m)/(0.24kg)*(0.54m^2-0.080^2)`

`v=√(6.54m^2/s^2)=2.556m/s`