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Hitman42 [59]
3 years ago
9

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?
Physics
1 answer:
bonufazy [111]3 years ago
5 0

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

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