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

`v_W=9.23m/s`

Step-by-step 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=(V^2)/(R)`

Dry pavement

`u_(KD)*g=(v_D^2)/(R)`

Wet pavement

`u_(KW)*g=(v_W^2)/(R)`

`u_(KW)=(1)/(3)*u_(KD)`

Solve and reduce the factor so:

`(v_W^2)/(v_D^2)=((1)/(3)*u_(KD))/(u_(KD))`

`v_W^2=v_D^2*(1)/(3)`

`v_W=v_D*(1)/(√(3))=16m/s*(1)/(√(3))`

`v_W=9.23m/s`