how do i get the max speed in m/sec for car rounding a curve of radius 10 m and coef of friction = 0.6?

Question

i need the formula and procedure, thank you :D

Answer 1

let mass of car = M

0.6 = Force of Friction / Normal Force , Normal Force = 9.81xM

Force of Friction = 5.886xM

this means that

centri force must not exceed 5.886xM

a=v^2 / 10, F_centri = (v^2 / 10)xM

(v^2 / 10)xM=5.886xM , M's cancel out

v=7.67

Answer 2

The formula: F=mv^2/r , where F is the centripetal force in Newton, m the mass of the car in kg, v is the speed we are looking for, and r is the radius of the curve in m.

The other formula: f=uR, where f is the force of friction in Newton, u is the coefficient of friction, and R is the force normal to the road. R is the weight of the car mg in this problem. We know that g=9.8m/s^2.

Now, in order that the car will not skid off the road, F should be equal to f. Therefore, substitute known values in our two formulas, and solve for v, as follows:

mv^2/10=0.60*mg

(m cancels out)

v^2=10*0.60*9.8
=58.8
v=7.67m/s

Answer 3

use f = m * a

there are 2 forces 1-friction 2-centerfution

for the car to stay in the coerce those 2 forces must be equal


m* v^2 / 10 = m*9.8 *.6 by tacking out " m " we get


v^2 = 5.88*10 = 58.8


v= 58.8 ^ (.5) = 7.67 m/s

Answer 4

equate centripetal force and friction force. m*v^2/r=mg*C_fric thats the point when it starts slipping. v=sqrt(9.8m/s^2*0.6*10m) =7.67m/s. Assuming flat (not inclined) road.