Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic

Question

Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g = 10 m/s²):

Accepted Answer

1. **Unit Conversion:** First, convert the initial speed from km/h to m/s.
   $$ u = 72 	ext{ km/h} = 72 	imes \frac{5}{18} 	ext{ m/s} = 20 	ext{ m/s} $$
2. **Deceleration due to Friction:** The retarding force is the kinetic friction, $f_k = \mu_k N = \mu_k mg$. According to Newton's Second Law, the deceleration $a$ is:
   $$ a = \frac{f_k}{m} = \frac{\mu_k mg}{m} = \mu_k g $$
   $$ a = 0.5 	imes 10 	ext{ m/s}^2 = 5 	ext{ m/s}^2 $$
   [Source 85, 87].
3. **Stopping Distance:** Using the kinematic equation $v^2 = u^2 - 2as$, where final velocity $v = 0$:
   $$ 0 = (20)^2 - 2(5)s $$
   $$ 400 = 10s $$
   $$ s = 40 	ext{ m} $$
   Alternatively, using the work-energy theorem: Work done by friction = Change in Kinetic Energy ($ \mu_k mg s = \frac{1}{2}mu^2 $), which yields the same result [Source 50, 98].

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