Question

A 1.450 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to displace the glider to a new equilibrium position, x= 0.250 m.

1. Find the effective spring constant of the system.

2. The glider is now released from rest at x= 0.250 m. Find the maximum x-acceleration of the glider.

3. Find the x-coordinate of the glider at time t= 0.610T, where T is the period of the oscillation.

4. Find the kinetic energy of the glider at x=0.00 m.

Answer 1

1. \(\mathrm{F}=\mathrm{kx}-\mathrm{r}>\mathrm{k}=\mathrm{F} / \mathrm{x}=0.9 / 0.25=3.6 \mathrm{~N} / \mathrm{m}\)

2. \(\mathrm{a}=\mathrm{kx}_{0} / \mathrm{m}=3.6^{*} 0.25 / 1.45=0.620689655 \mathrm{~m} / \mathrm{s}^{2}\)

3. \(x=x_{0} \cos (2 \pi t / T)\) \(-->x=0.25^{*} \cos \left(2^{*} 3.14159^{*} 0.61\right)=-0.192628827 \mathrm{~m}\)

4.kinetic energy \(=(1 / 2) \mathrm{k} \times 0^{2}=(1 / 2)^{*} 3.6^{*}\left(0.25^{\wedge} 2\right)=0.1125 \mathrm{~J}\)