A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking 30 . If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: (g = 10 m/s2, 3 = 1.73 )
- 51.4 m/s
- 102.8 m/s
- 70.4 m/s
- 264 m/s
Answer
(1) 51.4 m/s
Solution
[figure]
At v the car tends to slip up, so friction acts down the slope
Vertical: N = mg + N ; horizontal: N + N = mv2r
Divide: v = rg + 1 - (mass cancels)
30 = 1 3; multiply above and below by 3:
v = 3000 1 + 0.2 1.731.73 - 0.2 = 3000 1.3461.53
= 2639 51.4 m/s
Answer: (1)