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A 700 kg car makes a turn going at 30 m/s with radius of

curvature of 120 m. What is the force of friction between the car's tires and the road?

User Adva
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2 Answers

6 votes

Answer:5250 N

Explanation: ig:iihoop.vince

User Sivakumar
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4 votes

The force of friction between the car's tires and the road is 5250 N

Centripetal force is the force that is perpendicular to an object moving in circular motion. It is given by:

F = mv²/r

where m is the mass of object, v is the velocity and r is the radius of circle.

Given that m = 700 kg, v = 30 m/s, r = 120 m, hence:

F = mv²/r = 700 * 30²/120 = 5250 N

The force of friction between the car's tires and the road is 5250 N

User John Resig
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