WHen you take your 1100kg car out for a spin, you go around a corner of radius 55m with a speed of 15m/s. assuming your car doesnt skid, what is the force exerted on it by static friction?
The force has to equal centripetal force
mv^2/r
To find the force exerted on the car by static friction, we need to consider the centripetal force acting on the car as it goes around the corner.
Centripetal force can be calculated using the formula:
F = (m * v^2) / r
where:
F is the centripetal force,
m is the mass of the car,
v is the speed of the car, and
r is the radius of the corner.
Given:
m = 1100 kg (mass of the car)
v = 15 m/s (speed of the car)
r = 55 m (radius of the corner)
Plugging in the values into the formula, we get:
F = (1100 kg * (15 m/s)^2) / 55 m
Simplifying the equation:
F = (1100 kg * 225 m^2/s^2) / 55 m
F = (247,500 kg⋅m^2/s^2) / 55 m
F = 4,500 kg⋅m/s^2
The force exerted on the car by static friction is 4,500 kg⋅m/s^2.