Suppose a 1800-kg car passes over a bump in a roadway that follows the arc of a circle of radius 18.4 m as in the figure shown below.
(a) What force does the road exert on the car as the car passes the highest point of the bump if the car travels at 9.24 m/s? (Neglect any friction that may occur.)
magnitude
im getting 8352.a56N but it is incorrect i did 1800*9.24^2/18.4 and i got 835
Incorrect: Your answer is incorrect.
N
direction
Correct: Your answer is correct.
(b) What is the maximum speed the car can have without losing contact with the road as it passes this highest point?
13.44
Correct: Your answer is correct.
m/s
To solve part (a) of the problem, we need to use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration: F = m*a.
In this case, the net force is provided by the road, and we need to find the force exerted by the road on the car at the highest point of the bump. At this point, the car is in an accelerated circular motion, experiencing the centripetal force provided by the road.
The centripetal force is given by the equation F = 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 circular path.
So, to find the force exerted by the road on the car, we need to substitute the given values into the equation:
F = (1800 kg) * (9.24 m/s)^2 / (18.4 m)
Calculating this expression, we get:
F = 8352.446 N
Therefore, the correct answer for part (a) is 8352.446 N, not 835 N.
For part (b) of the problem, we need to find the maximum speed the car can have without losing contact with the road at the highest point. As the car goes over the bump, it needs to have sufficient centripetal force to counteract gravity and keep it in contact with the road.
At the highest point of the bump, the gravitational force acting on the car is equal to its weight, which is given by the equation Fg = m * g, where g is the acceleration due to gravity.
To find the maximum speed, we need to equate the gravitational force and the centripetal force:
Fg = F
m * g = (m * v^2) / r
Simplifying this equation, we get:
v^2 = g * r
Plugging in the values for g and r:
v^2 = (9.8 m/s^2) * (18.4 m)
Calculating this expression, we get:
v^2 = 180.32 m^2/s^2
Taking the square root of both sides of the equation, we get:
v = √(180.32) ≈ 13.44 m/s
Therefore, the correct answer for part (b) is 13.44 m/s.
v^2/R = 9.24^2/18.4
= 4.64 m/s^2 centripetal acceleration
so
apparent weight on road
= m (g-4.64)
= 1800 (9.81 - 4.64)
= 9306 N