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A police officer in hot pursuit drives her car through a circular turn of radius \(300 \mathrm{~m}\) with a constant speed of \(80.0 \mathrm{~km} / \mathrm{h} .\) Her mass is \(55.0 \mathrm{~kg}\). What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.

Short Answer

Expert verified
Net force magnitude: 546.62 N; Angle: 9.6°.

Step by step solution

01

Convert Speed to Meters per Second

To solve this problem, we first need to convert the speed from km/h to m/s. The speed is given as \(80.0\, \text{km/h}\). To convert to m/s, we use the conversion factor \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). This gives us: \[ 80.0 \times \frac{1}{3.6} = 22.22 \text{ m/s}\].
02

Calculate Centripetal Force

The centripetal force required to keep the officer moving in a circle can be calculated using the formula \( F_c = \frac{mv^2}{r}\), where \(m\) is the mass, \(v\) is velocity, and \(r\) is the radius. Plug in the values: \(m = 55.0 \text{ kg}\), \(v = 22.22 \text{ m/s}\), \(r = 300 \text{ m}\). Thus, \[ F_c = \frac{55.0 \times (22.22)^2}{300} = 90.74 \text{ N}\].
03

Calculate Gravitational Force

The gravitational force acting on the officer is due to her weight, calculated by \( F_g = mg\), where \(g = 9.8 \text{ m/s}^2\) is acceleration due to gravity. Thus, \[ F_g = 55.0 \times 9.8 = 539 \text{ N}\].
04

Determine the Net Force Magnitude

The net force on the officer is the vector sum of the centripetal force (horizontal) and the gravitational force (vertical). Use the Pythagorean theorem: \(F = \sqrt{F_c^2 + F_g^2}\). Substitute the values: \[ F = \sqrt{(90.74)^2 + (539)^2} = 546.62 \text{ N}\].
05

Calculate the Angle of the Net Force

The angle \(\theta\) relative to the vertical can be calculated using \(\tan\theta = \frac{F_c}{F_g}\). Solve for \(\theta\): \(\tan^{-1}\left(\frac{90.74}{539}\right)\). This gives us \(\theta = 9.6^\circ\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Centripetal Force is the force that keeps an object moving in a circular path. It acts towards the center of the circle and maintains the object's circular motion. This concept is crucial in understanding why cars don’t drift outwards while negotiating curves. When an officer drives through a circular turn, like in the problem, the car needs a certain force to keep moving along the curve. This is where centripetal force comes into play.

To determine this force, we use the formula: \[ F_c = \frac{mv^2}{r} \] Where:
  • \(m\) is the mass of the object (here, the police officer and her equipment together)
  • \(v\) is the velocity, converted into meters per second for accuracy.
  • \(r\) is the radius of the turn or circular path.
In our example, calculating with these values ensures the officer maintains her path without veering off. If either the mass, speed, or radius changes, it affects how much force is needed to keep her on track.
Gravitational Force
Gravitational Force is the force of attraction between two masses. On Earth, this force is experienced as weight and acts downwards towards the planet’s center. The force due to gravity can be calculated using the equation:\[ F_g = mg \]Here, \(m\) is the mass and \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\) on Earth.

For the police officer in the problem, the gravitational force acts vertically downward and contributes to the net force experienced while driving in a circle. Unlike centripetal force, which is horizontal and keeps the car moving in a circle, gravitational force pulls towards the ground, holding everything in the vehicle firmly against the car seats.

Understanding these forces not only helps in predicting how an object, like a car, will behave on Earth but also in everyday applications like determining safe speed limits on roads.
Vector Addition
Vector Addition involves combining different forces acting on an object to find one overall force, known as the resultant or net force. In situations where forces are not aligned in the same direction, vector addition is crucial to understanding the actual effect of multiple forces acting simultaneously.

In our exercise, both the centripetal and gravitational forces must be considered to determine the net force acting on the officer. These forces are perpendicular since the centripetal force acts horizontally while the gravitational force acts vertically downwards.

To combine them, we use the Pythagorean theorem because the forces are at a right angle to each other:\[ F = \sqrt{F_c^2 + F_g^2} \] This equation helps us find the magnitude of the net force. In vector terms, it is the "hypotenuse" of a right-angled triangle formed by these forces.

Understanding how to mathematically add vectors is essential in physics, helping predict movement and force interactions accurately, whether in complex machinery or simple daily life scenarios.

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Most popular questions from this chapter

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h} .\) Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m},\) what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

Two blocks, of weights \(3.6 \mathrm{~N}\) and \(7.2 \mathrm{~N},\) are connected by a massless string and slide down a \(30^{\circ}\) inclined plane. The coefficient of kinetic friction between the lighter block and the plane is \(0.10,\) and the coefficient between the heavier block and the plane is \(0.20 .\) Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string.

A baseball player with mass \(m=79 \mathrm{~kg},\) sliding into second base, is retarded by a frictional force of magnitude \(470 \mathrm{~N}\). What is the coefficient of kinetic friction \(\mu_{k}\) between the player and the ground?

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN},\) and the circle's radius is \(10 \mathrm{~m} .\) At the top of the circle, what are the (a) magnitude \(F_{B}\) and \((\mathrm{b})\) direction (up or down) of the force on the car from the boom if the car's speed is \(v=5.0 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{B}\) and (d) the direction if \(v=12 \mathrm{~m} / \mathrm{s} ?\)

A banked circular highway curve is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). The radius of the curve is \(200 \mathrm{~m}\). Traffic is moving along the highway at \(40 \mathrm{~km} / \mathrm{h}\) on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

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