91影视

• The radius of the circle,$\mathrm{R}=470\text{鈥尘}$.
• The horizontal component of force,${\mathrm{F}}_{\mathrm{h}}=210\text{鈥塏}$.
• The vertical component of force,${\mathrm{F}}_{\mathrm{v}}=500\text{鈥塏}$.
• Mass of the passenger, $\mathrm{m}=51\text{鈥�}\text{kg}$.

The railway car is moving around the horizontal circle at a constant speed. Thus, the horizontal component of the force must supply the centripetal force, which is necessary for maintaining the circular motion.

Formula:

$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

The upward force exerted by the car on the passenger is equal to the downward force of gravity$(\mathrm{W}=500\text{鈥�}\text{N})$on the passenger. So the net force does not have a vertical contribution; it only has the contribution from the horizontal force (which is necessary for maintaining the circular motion). Thus,

$\begin{array}{c}\left|{\stackrel{鈫�}{\mathrm{F}}}_{\mathrm{net}}\right|=\mathrm{F}\\ =210\text{N}\end{array}$

Thus, the magnitude of the net force on the passenger is 210 N.

Centripetal force is given by,

$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

$\mathrm{v}=\sqrt{\frac{\mathrm{FR}}{\mathrm{m}}}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{\frac{\left(210\text{鈥塏}\right)\left(470\text{鈥尘}\right)}{51.0\text{鈥塳驳}}}\\ \mathrm{v}=44.0\text{鈥尘/蝉}\end{array}$

Thus, the speed of the car is 44 m/s.