91影视

The "Sprong" obey the force law $F_x=-q{\left(x-x_{eq}\right)}^3$

Where;

$q=$the sprong constant.

$x_{eq}=$the equilibrium position.

$x_{eq}=0$

Thus; $F_x=-q{x}^3$

Mass of plastic ball$=20g$

The sprong constant $=40,000.$

The compression of sprong $=10cm$

For the unit of q,

Consider the equation,

$F_x=-q{x}^3$

$F_x$is in Newton (N)

x is in meters (m)

Hence;

$$\begin{gathered} F_x=-q{x}^3 \\ N=q{\left(m\right)}^3 \end{gathered}$$

for the unit, the sign does not matter

localid="1649412728684" $q=\frac{N}{m^3}$

The "Sprong" obey the force law $F_x=-q{\left(x-x_{eq}\right)}^3$

Where;

$q=$the sprong constant

$x_{eq}=$the equilibrium position.

$x_{eq}=0$

Thus; $F_x=-q{x}^3$

Mass of plastic ball$=20g$

The sprong constant$=40,000.$

The compression of sprong$=10cm$

The potential energy of a sprong:

"Sprong" obey the force law $F_x=-q{\left(x-x_{eq}\right)}^3$

Where;

$q=$the sprong constant

$x_{eq}=$the equilibrium position.

$x_{eq}=0$

Thus;$F_x=-q{x}^3$

Mass of plastic ball$=20g$

The sprong constant$=40,000.$

The compression of sprong$=10cm$

Applying the energy conservation equation to the ball and sprong system:

$$\begin{gathered} K_i+U_i=K_f+U_f \\ \frac{1}{2}m{v}_f^2+\left(0J\right)=\left(0J\right)+\frac{q{x}_i^4}{4} \\ 鈬�v_f=\sqrt{\frac{q{x}_i^4}{2m}} \\ {v}_f=\sqrt{\frac{q{x}_i^4}{2m}} \\ =\sqrt{\frac{\left(40,000\mathrm{N}/{\mathrm{m}}^3\right)(10\mathrm{cm}{)}^4}{\left(2\right)(20\mathrm{g})}} \\ =\sqrt{\frac{\left(40,000\mathrm{N}/{\mathrm{m}}^3\right)脳{\left(0.1\mathrm{m}\right)}^4}{2脳\left(\frac{20}{1000}\mathrm{Kg}\right)}} \\ =10m/s \end{gathered}$$