Question

1. a 1.2 × 10³ kg car rounds a curve of radius 50.0 m at a speed of 80.0 km/h (22 m/s).

(a) what is the centripetal acceleration of the car?
(b) how much centripetal force is needed to cause this acceleration?
(c) if the coefficient of kinetic friction, μ, is 0.25 on a slippery road, will the force of
friction between the road and the wheels of the car be enough to keep the car from
skidding?

1. a pendulum 1 00 - 1

Explanation:

Step1: Calculate centripetal acceleration

The formula for centripetal acceleration \(a_c=\frac{v^{2}}{r}\).
Given \(v = 22\ m/s\) and \(r=50.0\ m\).
\(a_c=\frac{22^{2}}{50}=\frac{484}{50}=9.68\ m/s^{2}\)

Step2: Calculate centripetal force

The formula for centripetal force \(F_c = ma_c\).
Given \(m = 1.2\times10^{3}\ kg\) and \(a_c=9.68\ m/s^{2}\).
\(F_c=(1.2\times 10^{3})\times9.68 = 1.1616\times10^{4}\ N\)

Step3: Compare friction force and centripetal force

The maximum friction force \(F_f=\mu_k N\). Since \(N = mg\) (normal force, \(m = 1.2\times10^{3}\ kg\), \(g = 9.8\ m/s^{2}\)), \(N=1.2\times10^{3}\times9.8 = 1.176\times10^{4}\ N\).
\(F_f=\mu_k N=0.25\times1.176\times 10^{4}=2.94\times10^{3}\ N\)
Since \(F_f(2.94\times 10^{3}\ N)<F_c(1.1616\times10^{4}\ N)\), the car will skid.

Answer:

(a) The centripetal acceleration is \(9.68\ m/s^{2}\).
(b) The centripetal force is \(1.16\times10^{4}\ N\).
(c) The car will skid.