A glider with mass $m = 2 \; kg$ sits on a frictionless horizontal air track, connected to a spring with force constant $k = 5 \; N/m$. You pull the glider, stretching the spring $0.1 \; m$ and then release it with no initial velocity. The glider begins to move back towards ie equilibrium position $(x = 0)$. What is its velocity when $x = 0.08 \; m$?

Given,

Mass $(m) = 2\;kg$

Force Constant $(k) = 5\;N/m$

Amplitude $(r) = 0.1\;m$

Position $(x) = 0.08\;m$

Velocity at $x = 0.08\;m$ = ?

We have,

$\frac{1}{2}m\;v^2 + \frac{1}{2}k\;x^2 = \frac{1}{2}k\;r^2$

By solving,

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

or, $v = \sqrt{\frac{5}{2}((0.1)^2 - (0.08)^2)}$

or $v = 0.09\;m/s$

∴ Velocity at $x = 0.08\;m$ is $0.09\;m/s$.

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