Answer to Question #96571 in Mechanics | Relativity for Ebube Paul Nwokeji

Question

A Glider with mass equals 0.2kg sits on a frictionless horizontal air track, connected to a spring of negligenble mass with force constant K=5.0N/M. You pull on the Glider stretching the spring 0.100m and then release it with no initial velocity. The Glider begins to move back towards its equilibrium position(S=0).
A) What is its speed when x=0.0800
B) What is the value of x when the object speed is 0.15m/s?

Answer 1

A) From the conservation of energy:

$$0.5mv^2+0.5kx^2=0.5kA^2$$

$$v=\sqrt{\frac{k}{m}(A^2-x^2)}$$

$$v=\sqrt{\frac{5}{0.2}(0.1^2-0.08^2)}=0.3\frac{m}{s}$$

B) From the conservation of energy:

$$0.5mv'^2+0.5kx'^2=0.5kA^2$$

$$x'=\sqrt{A^2-\frac{m}{k}v'^2}$$

$$x'=\sqrt{0.1^2-\frac{0.2}{5}0.15^2}=0.095\ m$$