Question

A driver takes a 1400-kg car out for a spin, going around a corner with a radius of 63 m at a speed of 18 m/s. The coefficient of static friction between the car and the road is 0.85. Assuming the car doesn’t skid, what is the force exerted on it by static friction?

Answer 1

Explanation:

Explanation:For Balancing The car static friction should balance the Centrifugal force ..

so F=mv2÷r

F=1400×18×18÷63

=7200..

Answer 2

Answer:

7200 N is the force exerted on the car by static friction

Explanation:

Given

Mass of a car (m)= 1400 kg

Radius of a road(r)= 63 m

Speed of the car(v)= 18 m/s

Coefficient of static friction = 0.85

To find

Force exerted by a static friction i.e centrifugal force

Explanation:

There are two types of friction come into phase when the car is in the motion along the curved path.They are

i)Static friction

ii)Kinetic friction

Here the car doesn't skid which implies that it supplies the net centrifugal force acting on the car. This force prevent the car from skidding .

Step by step solution:

Force of static friction= Coefficient of static friction × Normal force

Where N = mg

F = mv^2 ÷ r

The above formula represents the centrifugal force

F =1400 × 18^ 2/63

F = 453600÷ 63

F= 7200 N

That implies 7200 N is provided by the centrifugal force as a static friction.

To know more about static friction, click on the link below

https://brainly.in/question/229980

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