Question

A 1900 kg car moves along a horizontal road at speed v0 = 17.3 m/s. The road is wet, so the static friction coefficient between the tires and the road is only µs = 0.136 and the kinetic friction coefficient is even lower, µk = 0.0952. The acceleration of gravity is 9.8 m/s2 . Assume: No aerodynamic forces; g = 9.8 m/s2, forward is the positive direction. What is the highest possible deceleration of the car under such conditions? Answer in units of m/s2.

Answer 1

Answer on Question 63485, Physics, Mechanics, Relativity

Question:

A $1900 \, kg$ car moves along a horizontal road at speed $v_0 = 17.3 \, m/s$. The road is wet, so the static friction coefficient between the tires and the road is only $\mu_s = 0.136$ and the kinetic friction coefficient is even lower, $\mu_k = 0.0952$. The acceleration of gravity is $9.8 \, m/s^2$. Assume: No aerodynamic forces; $g = 9.8 \, m/s^2$, forward is the positive direction. What is the highest possible deceleration of the car under such conditions? Answer in units of $m/s^2$.

Solution:

We can find the highest possible deceleration of the car under such conditions from the Newton's Second Law of Motion:

$$F_{net} = m a_{max},$$

here, $F_{net}$ is the net force that acts on the car, $m$ is the mass of the car and $a_{max}$ is the highest possible deceleration of the car.

The only net force that acts on the car is the force of friction, so we can write:

$$F_{fr} = m a_{max},$$ $$\mu_k m g = m a_{max},$$ $$a_{max} = \mu_k g = 0.0952 \cdot 9.8 \, \frac{m}{s^2} = 0.93 \, \frac{m}{s^2}.$$

Answer:

$$a_{max} = 0.93 \, \frac{m}{s^2}.$$

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