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

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

$x_2=0.08m$

$v_1=0 m/s$

Solve to velocity v2

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

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

$v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)$

$v=\sqrt{6.54m^2/s^2}=2.556m/s$