Question

find the centripetal force needed by a 1275 kg car to make a turn of radius 40.0 m at a speed of 25.0 km/h. 1540 n 797 n 221 n 1.99×10^{4}n

Explanation:

Step1: Convertir la velocidad a m/s

$$v = 25.0\frac{km}{h}\times\frac{1000m}{1km}\times\frac{1h}{3600s}\approx6.94m/s$$

Step2: Aplicar la fórmula de fuerza centrípeta

La fórmula de fuerza centrípeta es $F_c=\frac{mv^{2}}{r}$, donde $m = 1275kg$, $v\approx6.94m/s$ y $r = 40.0m$.
$$F_c=\frac{1275kg\times(6.94m/s)^{2}}{40.0m}$$
$$F_c=\frac{1275kg\times48.16m^{2}/s^{2}}{40.0m}$$
$$F_c=\frac{61403.4kg\cdot m^{2}/s^{2}}{40.0m}$$
$$F_c = 1535.085N\approx1540N$$

Answer:

1540 N