Given:
Construct a force diagram for the car when it's at the top of the hill. (Justify the relative forces in your force diagram.)
This force diagram includes the weight force that the Earth exerts downward on the car, as well as the normal force that the road exerts upward on the car. We know that these are equal since the car does not fall through the ground nor float away from it. The force diagram also includes the friction force which moves the car to the right. This is caused by the tires rubbing against the road. It pushes the car forward since the tires move clockwise and push against the road because friction opposes motion.
Supposed the speed of the car is 11.1m/s (25mph), and the radius of curvature (r) is 25m; determine the magnitude of the centripetal acceleration of the car.
a = v^2 / r
a = (11.1m/s)^2 / 25m
a = 123.2m^2/s^2 / 25m
a = 4.928 m/s^2
To find acceleration, we use the forula acceleration equals velocity squared divided by radius. Since we are given both velocity (11.1m/s) and radius (25m), we simply plug them in to the equation and solve. The answer is 4.928m/s^2. This means that the car's velocity increases 4.928m/s as each second goes by.
ΣF = m * a
ΣF = 1200kg * 4.928m/s^2
ΣF = 5913.6N
Since we are given the car's mass and have already found its acceleration, we can find net force by using the equation net force equals mass times acceleration. When we plug them in, we find that net force equals 5913.6N. This means that 5913.6N of force would be necessary in order to cause the car's centripetal acceleration of 4.928m/s^2.
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