A 975-kg car (including the driver) crosses the rounded top of a hill of radius 88...

A 975-kg car (including the driver) crosses the rounded top of a hill of radius 88 m at a speed of 12 m.s-1.

Calculate: (i) the normal force exerted on the car by the road

(ii) the normal force exerted by the car on the 72-kg driver

(iii) the sped of the car at which the normal force on the driver is zero.

Solutions

Expert Solution

Given the mass of the car and driver togeher is M = 975kg, the radius of the cirular track r = 88m and the speed of the car is v = 12m/s.

(i) When the car is in the rounded top of the hill, the weight of the car (W = Mg) will act downwards and the normal force N exerted by the road on the car will act upwards. The resultant of these forces will give the necessary centripetal force. Therefore,

So the normal force exerted on the car by the road is 7959.55N.

(ii) Given the weight of the driver is m = 72kg. Since the driver is inside the car, the velocity of the driver and car is the is the same. For the driver, his weight (W = mg) will act downwards and the normal force N exerted by the car on the driver will act upwards. The resultant of these forces will give the necessary centripetal force. Therefore,

So the normal force exerted by the car on the driver is 587.78N.

(iii) When the normal force is zero, the eqation of the car becomes,

The speed of the car when the normal force is zero is 29.37m/s.

genius_generous answered 1 year ago

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