A high-speed railway car goes around a flat, horizontal circle of radius 470mat a constant speed. The magnitudes of the horizontal and vertical components of the force of the car on a 51.0kgpassenger are 210Nand 500N, respectively.

(a)What is the magnitude of the net force (of allthe forces) on the passenger?

(b) What is the speed of the car?

• The radius of the circle, R =470m.
• The horizontal component of force, F_h = 210N.
• The vertical component of force, F_v = 500N.
• Mass of the passenger, m = 51kg.

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:

$F=\frac{m{v}^2}{R}$

The upward force exerted by the car on the passenger is equal to the downward force of gravity (W = 500 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,

$\begin{array}{rcl}\left|{\stackrel{\rightharpoonup }{F}}_{net}\right|& =& F\\ & =& 210N\end{array}$

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

Centripetal force is given by,

$$\begin{gathered} F=\frac{m{v}^2}{R} \\ v=\sqrt{\frac{FR}{m}} \end{gathered}$$

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

$$\begin{gathered} v=\sqrt{\frac{\left(210N\right)\left(470m\right)}{51.0kg}} \\ v=44.0m/s \end{gathered}$$

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