A police officer in hot pursuit drives her car through a circular turn of radius 320 m with a constant speed of 74.5 km/h. Her mass is 55.0 kg.
(a) What is the magnitude of the net force of the officer on the car seat?
______N
(b) What is the angle (relative to vertical) of the net force?
_________°
mg is downward, mv^2/r will be horizontal, directed inward to the turn.
To solve these questions, we need to consider the forces acting on the police officer and the car as it goes through the circular turn.
First, let's find the acceleration of the car as it moves through the turn. We can use the centripetal acceleration formula:
a = v^2 / r
where:
a = acceleration
v = velocity
r = radius
Given:
v = 74.5 km/h * (1000 m/1 km) * (1/3600 h/1s) = 20.7 m/s
r = 320 m
Plugging in the values into the formula:
a = (20.7 m/s)^2 / 320 m = 1.34 m/s^2
Now, let's calculate the net force acting on the officer. The net force is equal to the mass of the officer multiplied by the acceleration:
Fnet = m * a
Given:
m = 55.0 kg
a = 1.34 m/s^2
Plugging in the values into the formula:
Fnet = (55.0 kg) * (1.34 m/s^2) = 73.7 N
(a) The magnitude of the net force of the officer on the car seat is 73.7 N.
To find the angle of the net force relative to the vertical, we need to consider the vertical and horizontal components of the net force. Since the net force is directed towards the center of the circular turn, the vertical and horizontal components must be perpendicular.
The vertical component of the net force is equal to the weight of the officer:
Fvertical = m * g
where:
m = mass of the officer
g = acceleration due to gravity ≈ 9.8 m/s^2
Given:
m = 55.0 kg
Plugging in the values into the formula:
Fvertical = (55.0 kg) * (9.8 m/s^2) = 539 N
The horizontal component of the net force is equal to the centripetal force:
Fhorizontal = Fnet
Given:
Fnet = 73.7 N
Therefore, the angle between the net force and the vertical is given by:
tan(angle) = Fhorizontal / Fvertical
angle = arctan(Fhorizontal / Fvertical)
Plugging in the values:
angle = arctan(73.7 N / 539 N)
Using a calculator, we find that the angle is approximately 7.8°.
(b) The angle (relative to vertical) of the net force is approximately 7.8°.