A car of weight 20 kN moves at constant speed on a straight but hilly road. One section has a crest and dip of the same 90 m radius. (a) As the car passes over the crest the normal force is one half t

Question

A car of weight 20 kN moves at constant speed on a straight but hilly road. One section has a crest and dip of the same 90 m radius. (a) As the car passes over the crest the normal force is one half the weight of the car. What will be the normal force as it passes through the bottom of the dip? (b) What is the greatest speed at which the car can move without leaving the road at the top of the hill? (c) Moving at the speed found in part (b), what will be the normal force at the bottom of the dip? (Take $g = 10 m / s^{2}$ )

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Answer:

A

Solution:

(a) $\frac{m v ^{2}}{R} = 20 - 10 = 10 kN$ ,
so $N_{2} = 20 + 10 = 30 kN$ .
(b) $v_{max} = g R$
$= 900$
$= 30 m / s$ .
(c) At $v_{max}, \frac{m v ^{2}}{R} = m g$
$= 20 kN$ ,
so $N = 20 + 20 = 40 kN$ .

TestHub · Accepted Answer

(a) R m v 2 ​ = 20 − 10 = 10 kN , so N 2 ​ = 20 + 10 = 30 kN . (b) v max ​ = g R ​ = 900 ​ = 30 m / s . (c) At v max ​ , R m v 2 ​ = m g = 20 kN , so N = 20 + 20 = 40 kN .

Physics - Circular Motion Question with Solution | TestHub

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