contestada

Wings on race cars push them into the track. The increased normal force makes large friction forces possible. At one Formula One racetrack, cars turn around a half-circle with diameter 190m at 68m/s .
Part A
For a 610 kg vehicle, the approximate minimum static friction force to complete this turn is. Select best answer
a.15000N
b.6000N
c.30000N
d.18000N
e.24000N

Respuesta :

Answer:

a) F = 15000 N

Explanation:

At flat race track the friction force between the tyre and the road will provide the required frictional force in the form of centripetal force

So we will have

[tex]F_f = \frac{mv^2}{R}[/tex]

so here we know that

m = 610 kg

v = 68 m/s

R = 190 m

now we have

[tex]F = \frac{610 (68^2)}{190}[/tex]

[tex]F = 14845.5 N[/tex]

this is nearly

[tex]F = 15000 N[/tex]

ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE

Otras preguntas