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A 1500 kg car travels around a circular track at a constant speed. a) If the radius of the path is 200 m and the car is traveling at 45 m/s, how long does it take the car to do a lap around the track? b) What net force is acting on the car in the horizontal direction? c). What force keeps the car in a circular path? d). What is the coefficient of static friction between the car’s tires and the road?



Answer :

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a) The time taken is 4.44 s

b) The force that keeps it moving is 15188 N

c) The centripetal force is responsible for the motion in the circular path.

d) The coefficient of the friction is 1.03

What is the centripetal force?

The centripetal force is the force that keep the object moving in a circular path. In this case, we need to find the how long does it take the car to do a lap around the track.

a) The time that is taken is the radio of the distance to the speed of the object. Thus the time that is taken =  200 m/45 m/s = 4.44 s

b) The net force that acts on the car as it is negotiating the circular path is given as; mv^2/r = 1500 * (45)^2/200

= 15188 N

c) The centripetal force keeps the car on the circular path

d) The coefficient of static friction is; 15188 N/1500 kg * 9.8 m/s^2

= 1.033

Learn more about centripetal force:https://brainly.com/question/11324711

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