Question

A spring with a spring constant of 6.2 N m−1 is used to launch a 5.0 g slider along a frictionless track as shown in the figure above. The spring is compressed by 16 cm.


To what height h above the height labeled 0 m in the figure does the slider reach? (in m to 2 s.f)

Answer 1

$\frac{k\Delta x^2}{2}=\frac{mv^2}{2}\to v=\Delta x\sqrt{\frac{k}{m}}=0.16\sqrt{\frac{6.2}{0.005}}=5.63m/s$

$h=\frac{v^2}{2g}=\frac{5.63^2}{2\cdot9.81}\approx1.62m$