23P3Q08

8ai)

$\displaystyle k=\frac{F}{e}=\frac{4.00}{0.200}=20.0\text{ N }{{\text{m}}^{-1}}$

8aii)

$\displaystyle EPE=\frac{1}{2}Fe=\frac{1}{2}(3.0)(0.15)=0.225\text{ J}$

8aiii)

$\displaystyle \Delta GPE=mg\Delta h=(3)(0.15)=0.45\text{ J}$

8b)

In order for the object to not go into oscillation, an upward external force must have supported the object as it is lowered.

The difference corresponds to the negative work done by this external force.

8c)

The magnitude of the restoring force is kx,

and it is directed towards the equilibrium position.

By newton’s second law:

$\displaystyle \begin{aligned} & ({{F}_{net}}=ma) \\ & -kx=ma \\ & a=-\frac{k}{m}x \end{aligned}$

8di)

$\displaystyle \begin{aligned} \omega &=\sqrt{\frac{k}{m}} \\ &=\sqrt{\frac{20.0}{3\div 9.81}} \\ &=8.087\text{ rad }{{\text{s}}^{-1}} \end{aligned}$

$\displaystyle \begin{aligned} (x&={{x}_{0}}\sin \omega t) \\ x&=3.2\sin (8.087\times 0.50) \\ &=-2.51\text{ cm} \end{aligned}$

COMMENT: Did you set your calculator to radian mode?

COMMENT: The negative sign means that object has swung to the other side.

8dii)

Method 1

$\displaystyle \begin{aligned} (v&={{v}_{\max }}\cos \omega t) \\ v&=(8.087)(3.2)\cos (8.087\times 0.50) \\ &=-16.0\text{ cm }{{\text{s}}^{-1}} \end{aligned}$

Method 2

$\displaystyle \begin{aligned} v&=\pm \omega \sqrt{{{x}_{0}}^{2}-{{x}^{2}}} \\ &=\pm 8.087\sqrt{{{3.2}^{2}}-{{(-2.51)}^{2}}} \\ &=\pm 16.1\text{ cm }{{\text{s}}^{-1}} \end{aligned}$