A car of mass 1000 kg travels around a level curve of radius 40 m. If the maximum frictional force that can be exerted upon the car by the road (determined by the coefficient of friction between the tires and the road) is 7000 N how fast can the car travel without "spinning out?"

The car can travel V=19.8m/s without spinning out

Explanation:

Given that

Mass of the car m=1000kg

Radius of curve r =40m

frictional force R=7000N

Coefficient of friction u=?

We know that F=uR

But F=ma

F=1000*9.81

F=9810N

Therefore u=F/R

u=9810/7000

u=1.40

For the car to travel without spinning out we

Equate it centripetal force formula to frictional force formula

F=mv²/r

F=ur

hence mv²/r=uR

Making velocity subject of formula we have v²=u*r*R/m

V²=(1.4*40*7000)/1000

V²=392

V=√ 392

V=19.8m/s

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

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