Question

Concept Simulation 5.2 reviews the concepts that are involved in this problem. A car is safely negotiating an unbanked circular turn at a speed of 16 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

Answer 1

Answer:

$v_W=9.23m/s$

Explanation:

The force of friction change is the pavement is dry or wet so to determine the force of friction:

$F=m*a$

$F_k=m*a_c$

$F_k=u_K*N$

$N=m*g$

$F_k=u_K*m*g=m*a_c$

$u_K*g=a_c$

$a_c=\frac{V^2}{R}$

Dry pavement

$u_{KD}*g=\frac{v_D^2}{R}$

Wet pavement

$u_{KW}*g=\frac{v_W^2}{R}$

$u_{KW}=\frac{1}{3}*u_{KD}$

Solve and reduce the factor so:

$\frac{v_W^2}{v_D^2}=\frac{\frac{1}{3}*u_{KD}}{u_{KD}}$

$v_W^2=v_D^2*\frac{1}{3}$

$v_W=v_D*\frac{1}{\sqrt{3}}=16m/s*\frac{1}{\sqrt{3}}$

$v_W=9.23m/s$