A sports car of mass 950 kg (including the driver) crosses the rounded top of a hill (r = 88 m) at 27 m/s.
Determine the normal force exerted by the road on the car. Determine the normal force exerted by the car on the 72-kg driver.
Express your answer using two significant figures.
sorry i made some typing errors
B)=(72 * 9.8) - (72*729/88)
which i got 109 and that = 1.1 *10^2
C)v= sqrt(88*9.8)
the 91 was from my problem i had to do.
hope this helps
force =mg-mv^2/r where m is 72kg.
Is the equation above used to determine the normal force exerted by the road on the car?
When I used the equation I got (72)(9.8) - (72)(729)/88 = 1.2 (2 sig figs)
This answer was incorrect. What am I doing wrong.
And.... what equation can I use to figure out the 2nd part?
if you haven't already figured it out the equation is for both the driver and car,
a)you pluged in for driver but it should be weight of the car so its
=(950 * 9.8)-(950*729/88)
B)you have already figured out
=(72)(9.8) - (72)(729)/8
C) to determine the speed of the car with no normal force exerted on the driver you simply V= sqrt(91*9.8)
B=72-9-8
To determine the normal force exerted by the road on the car, we need to consider the forces acting on the car at the top of the hill.
First, let's calculate the gravitational force acting on the car at the top of the hill. The gravitational force is given by 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 = 950 kg * 9.8 m/s^2 = 9310 N
Since the car is on top of the hill, there is also a centrifugal force acting outward. The centrifugal force is given by the formula:
F_centripetal = m * (v^2 / r)
where v is the velocity of the car and r is the radius of the hill.
F_centripetal = 950 kg * (27 m/s)^2 / 88 m
F_centripetal = 7378 N
At the top of the hill, the normal force exerted by the road on the car is equal to the sum of the gravitational force and the centrifugal force:
F_normal = F_gravity + F_centripetal
F_normal = 9310 N + 7378 N
F_normal = 16690 N
Therefore, the normal force exerted by the road on the car is approximately 16690 N.
To determine the normal force exerted by the car on the driver, we can apply Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Since the normal force exerted by the road on the car is directed upward, the normal force exerted by the car on the driver is directed downward and has the same magnitude.
Hence, the normal force exerted by the car on the 72-kg driver is also approximately 16690 N.