F=m*((v^2)/r)
Therefore (18.1^2m/s)/80m*3800=F
Therefore F=15561N
a) what is the force that the road exerts on the car to keep it in motion around the corner?
b) what force would the road need to exert if the car was traveling at 100 km/hr?
Therefore (18.1^2m/s)/80m*3800=F
Therefore F=15561N
a) To calculate the force exerted by the road on the car at 65 km/hr, we can use the centripetal force formula, which is given by:
F = (m * v^2) / r
where,
F is the centripetal force,
m is the mass of the car (3800 kg),
v is the velocity of the car (converted to m/s),
r is the radius of curvature (80 m).
First, let's convert the velocity from km/hr to m/s:
65 km/hr * (1000 m / 3600 s) = 18.06 m/s (rounded to two decimal places)
Now, we can substitute the values into the formula:
F = (3800 kg * (18.06 m/s)^2) / 80 m
F = (3800 kg * 326.5236 m^2/s^2) / 80 m
F ≈ 15588 N (rounded to the nearest whole number)
Therefore, the force that the road exerts on the car to keep it in motion around the corner at 65 km/hr is approximately 15588 Newtons.
b) To calculate the force required if the car was traveling at 100 km/hr, we will use the same centripetal force formula, but with the new velocity:
100 km/hr * (1000 m / 3600 s) = 27.78 m/s (rounded to two decimal places)
Substituting this value for velocity into the formula:
F = (3800 kg * (27.78 m/s)^2) / 80 m
F = (3800 kg * 771.0484 m^2/s^2) / 80 m
F ≈ 14627 N (rounded to the nearest whole number)
Therefore, the force that the road would need to exert on the car to keep it in motion around the corner at 100 km/hr is approximately 14627 Newtons.