An 893 kg car passes over a bump in a road that follows the arc of a circle of radius 14.3 meters.
A) What force does the road exert on the car as the car passes the highest point of the bump if the car travels 7.53 m/s.
B) What is the max speed the car can have as it passes the highest point before losing contact with the road?
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To find the force exerted by the road on the car as it passes the highest point of the bump, we need to consider the gravitational force acting on the car and the normal force exerted by the road.
A) The gravitational force acting on the car can be calculated using the formula:
F_gravity = m * g
where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 893 kg * 9.8 m/s^2 = 8751.4 N
At the highest point of the bump, the normal force exerted by the road is equal in magnitude but opposite in direction to the gravitational force. So, the force exerted by the road can be calculated as:
F_road = - F_gravity
F_road = -8751.4 N
Note: The negative sign indicates that the force is acting downwards.
B) To find the maximum speed the car can have as it passes the highest point before losing contact with the road, we need to consider the net force acting on the car. At the highest point, the net force should be equal to zero to ensure that there is no contact force exerted by the road.
The net force can be calculated using the centripetal force formula:
F_net = m * v^2 / r
where m is the mass of the car, v is the velocity of the car, and r is the radius of the circle.
At the highest point, the net force is zero, so we can set the equation equal to zero:
0 = m * v^2 / r
Solving for v, the maximum speed, we have:
v^2 = 0
v = 0 m/s
This means that the maximum speed the car can have at the highest point before losing contact with the road is zero. Beyond this point, the car will lose contact and start to fall due to the force of gravity.