Question

A car turns a corner on a slippery road at a constant speed 10 m/s If the coefficient of friction is 0.5,the minimum radius of arc in meter in which the car turns..

Answer 1

Condition is, A car turns a corner on a slippery road , $v\leq\sqrt{\mu rg}$

where v is the speed of car , $\mu$ is coefficient of friction , r is the radius of arc in which the car turns and g is acceleration due to gravity.

now, v ≤ $\sqrt{\mu rg}$

squaring both sides,

v² ≤ $\mu rg$

v²/$\mu$g ≤ r

hence, minimum value of r = v²/$\mu$g

now, r = (10)²/(0.5 × 10) = 100/5 = 20m

hence, radius of arc in which the car turns is 20m

Answer 2

Answer:

20N

Explanation:

hope this will help you very much.