Answer to Question #109677 in Mechanics | Relativity for Shaima

Question

A glider with mass sits on a frictionless horizontal air track, connected to a
spring with force constant k = 5.00 N/m. You pull on the glider, stretching
the spring 0.100 m, and release it from rest. The glider moves back toward
its equilibrium position (x=0). What is its x-velocity when x = 0.080 m?

Answer 1

From the conservation of energy:

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

$$(0.2)v^2=(5)(0.1^2-0.08^2)$$

$$v=0.3\frac{m}{s}$$