A car of mass 720 kg travels round a horizontal, semicircular racetrack of radius 200 m, at a constant speed of 48 km/h. What is the sideways component of the force exerted by the track on the car’s tyres?
F = m v^2/r
where
m = 720
v = 48000/3600 = 13.33 m/s
r = 200
To find the sideways component of the force exerted by the track on the car's tires, we need to consider the centripetal force acting on the car as it moves in a circular path.
The centripetal force is given by the equation:
F = (m * v^2) / r
Where:
F is the force,
m is the mass of the car,
v is the velocity of the car, and
r is the radius of the circular path.
First, let's convert the velocity from kilometers per hour to meters per second:
48 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 13.33 m/s
Now, we can substitute the values into the equation:
F = (720 kg * (13.33 m/s) ^ 2) / 200 m
Simplifying the equation further:
F = (720 kg * 177.57 m^2/s^2) / 200 m
F = 6,791.04 kg⋅m/s^2
The unit kg⋅m/s^2 represents a force, which is equal to a Newton (N). So, the sideways component of the force exerted by the track on the car's tires is approximately 6,791 N.