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Q10.

Expert-verifiedFound in: Page 130

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**A car maintains a constant speed v as it traverses the hill and valley shown in Fig. 5–34. Both the hill and valley have a radius of curvature R. At which point, A, B, or C, is the normal force acting on the car (a) the largest, (b) the smallest? Explain. (c) Where would the driver feel heaviest and (d) lightest? Explain. (e) How fast can the car go without losing contact with the road at A?**

The normal force will be largest and smallest at point C and A, the driver feel the heaviest and lightest at point C and A, and the fastest speed of the car is $v=\sqrt{gR}$.

The constant speed of the car is $v$.

The radius of curvature is $R$.

**In this problem, for determining the largest and smallest normal forces exerting on the car, the centripetal force should be directed upwards.**

The normal force will be largest at point C because the direction of centripetal force is upwards and to provide the net upward force the normal force should be more than the weight of the car.

The normal force exerting on the car is lowest at the topmost point of the hill, point A. The centripetal force in this case is directed towards the centre that is downwards, hence the weight of car will be more than the normal force.

At the point C, the driver of the car will feel the heaviest because as mentioned in the above part the normal force is highest at this point.

At the point A, the driver of the car will feel the lightest because as mentioned in the above part the normal force is lowest at this point.

The maximum speed of the car without losing contact with the road is achieved at the point where the car almost loses the contact with the road related to the normal force. In this condition, the normal force will be zero.

The relation of vertical forces acting on the car is given by,

$\begin{array}{c}F=ma\\ mg-N=m\left(\frac{{v}^{2}}{R}\right)\end{array}$

On plugging the values in the above relation.

$\begin{array}{c}mg-0=m\left(\frac{{v}^{2}}{R}\right)\\ v=\sqrt{gR}\end{array}$

Thus, $v=\sqrt{gR}$ is the fastest speed of the car.

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