91影视

In an isolated system, the total of gravitational potential energy and kinetic energy is always constant, according to the rule of conservation of energy.

The skateboarder intends to move on a track with a quarter-circle radius, as shown in the diagram below. Point A is the skateboarder's beginning point, while point B is the skateboarder's highest point$\mathrm{B}$. The point's elevation$\mathrm{B}\left(y_{\mathrm{B}}\right)$is equal to the circular path's radius.

Consider the ball's movement from the projection point A to the highest point B. The total of gravitational potential energy and kinetic energy at these two sites is identical due to the rule of conservation of energy.

$U_{\mathrm{A}}+K_{\mathrm{A}}=U_{\mathrm{B}}+K_{\mathrm{B}}$

We have, $U_{\mathrm{A}}$is potential gravitational energy at the point $\mathrm{A},K_{\mathrm{A}}$is kinetic energy at the point $\mathrm{A}$, $U_{\mathrm{B}}$is gravitational potential energy at the point $\mathrm{B}$, and $K_{\mathrm{B}}$is kinetic energy at the point B.

Put $mg{y}_{\mathrm{A}}$for $U_{\mathrm{A}},mg{y}_{\mathrm{B}}$for $U_{\mathrm{B}},\frac{1}{2}m{v}_{\mathrm{A}}^2$for $K_{\mathrm{A}}$and $\frac{1}{2}m{v}_{\mathrm{B}}^2$for $K_{\mathrm{B}}$in the equation

$U_{\mathrm{A}}+K_{\mathrm{A}}=U_{\mathrm{B}}+K_{\mathrm{B}}.$

$mg{y}_{\mathrm{A}}+\frac{1}{2}m{v}_{\mathrm{A}}^2=mg{y}_{\mathrm{B}}+\frac{1}{2}m{v}_{\mathrm{B}}^2$

we have, $y_{\mathrm{A}}$is height of point $\mathrm{A},y_{\mathrm{B}}$is height of point $\mathrm{B},v_{\mathrm{A}}$is velocity at the point $\mathrm{A},v_{\mathrm{B}}$is velocity at the point B, $m$is the skateboarder's mass, and $g$gravitational acceleration is a term used to describe the acceleration caused by gravity.

Repeating the equation again $mg{y}_{\mathrm{A}}+\frac{1}{2}m{v}_{\mathrm{A}}^2=mg{y}_{\mathrm{B}}+\frac{1}{2}m{v}_{\mathrm{B}}^2$for $v_{\mathrm{A}}$.

$mg{y}_{\mathrm{A}}+\frac{1}{2}m{v}_{\mathrm{A}}^2=mg{y}_{\mathrm{B}}+\frac{1}{2}m{v}_{\mathrm{B}}^2$

$g{y}_{\mathrm{A}}+\frac{1}{2}{v}_{\mathrm{A}}^2=g{y}_{\mathrm{B}}+\frac{1}{2}{v}_{\mathrm{B}}^2$

$2g{y}_{\mathrm{A}}+v_{\mathrm{A}}^2=2g{y}_{\mathrm{B}}+v_{\mathrm{B}}^2$

$v_{\mathrm{A}}^2=2g{y}_{\mathrm{B}}-2g{y}_{\mathrm{A}}+v_{\mathrm{B}}^2$

$v_{\mathrm{A}}=\sqrt{2g\left(y_{\mathrm{B}}-y_{\mathrm{A}}\right)+v_{\mathrm{B}}^2}$

Putting equation$y_{\mathrm{B}}$with $R$and 0 for $y_{\mathrm{A}}$in the above equation.

$v_{\mathrm{A}}=\sqrt{2g(R-0)+v_{\mathrm{B}}^2}$

$=\sqrt{2gR+v_{\mathrm{B}}^2}$

Putting$3.0\mathrm{m}$for $R,0\mathrm{m}/\mathrm{s}$for $v_{\mathrm{B}}$, and $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$in the above equation.

$v_{\mathrm{A}}=\sqrt{2gR+v_{\mathrm{B}}^2}$

$=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(3.0\mathrm{m})-(0\mathrm{m}/\mathrm{s}{)}^2}$

$=7.7\mathrm{m}/\mathrm{s}$

As a result, the required minimum speed at the bottom of the track is $7.7\mathrm{m}/\mathrm{s}$.