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Chapter 9: Problem 69

A car with a mass of \(1000 .\) kg goes over a hill at a constant speed of \(60.0 \mathrm{~m} / \mathrm{s}\). The top of the hill can be approximated as an arc length of a circle with a radius of curvature of \(370 . \mathrm{m} .\) What force does the car exert on the hill as it passes over the top?

Short Answer

Expert verified
The force exerted by the car on the hill as it passes over the top is 80.27 N upwards.

Step by step solution

01

Identify given values

The mass of the car (m) is given as \(1000\, kg\). The constant speed (v) the car is moving at is \(60.0\, m/s\). The radius of curvature (R) for the top of the hill is given as \(370\, m\).
02

Calculate gravitational force

To find the gravitational force (F_g) acting on the car, we can use the formula \[F_g = m \times g\] where g is the acceleration due to gravity, approximately equal to \(9.81\, m/s^2\). So, \[F_g = (1000\, kg) \times (9.81\, m/s^2) = 9810\, N\].
03

Calculate centripetal force

To find the centripetal force (F_c) required to keep the car moving in its curved path over the top of the hill, we can use the formula \[F_c = \dfrac{m \times v^2}{R}\] where m is the mass of the car, v is its constant speed and R is the radius of curvature. Plugging in given values, \[F_c = \dfrac{(1000\, kg) \times (60.0\, m/s)^2}{370\, m} = 9729.73\, N\]
04

Determine normal force exerted by the hill

As the car is at the top of the hill, the centripetal force and gravitational force should have the same direction (downward). Therefore, to calculate the net force (F_net) of these two forces, we can use the formula: \[ F_{net} = F_c - F_g \] Plugging in the values, \[F_{net} = 9729.73\, N - 9810\, N = -80.27\, N\] The net force acting on the car is downwards and has a magnitude of \(80.27\, N\). Therefore, the normal force exerted by the hill on the car (F_n) is equal in magnitude and opposite in direction.
05

Calculate the force exerted by the car

The force exerted by the car on the hill (F_car) is equal in magnitude and opposite in direction to the normal force exerted by the hill on the car. Thus, \[F_{car} = -F_n = 80.27\, N\] So the force exerted by the car on the hill as it passes over the top is \(80.27\, N\) upwards.

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Key Concepts

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

Gravitational Force
Understanding gravitational force is essential when studying the movement of objects on Earth. It is the force that attracts a body towards the center of the planet, and its standard value鈥攔epresented by the letter 'g'鈥攊s approximately \(9.81 \text{ m/s}^2\).

For any object with mass, such as the car in our problem, the gravitational force \(F_g\) acting on it can be calculated with the equation \(F_g = m \times g\), where 'm' is the mass of the object. In our case, when a \(1000\text{ kg}\) car goes over a hill, the gravitational force exerted on it is \(9810\text{ N}\). This force plays a pivotal role in determining the car's interaction with the hill, particularly when combined with other forces like the centripetal force during circular motion.
Circular Motion
When a car travels in a path that forms an arc, such as over the round top of a hill, it is undergoing circular motion. This motion requires a centripetal force \(F_c\) to keep the car moving in a curved path rather than in a straight line.

The formula for centripetal force is \(F_c = \dfrac{m \times v^2}{R}\), where 'm' is the object's mass, 'v' is the velocity at which it鈥檚 moving, and 'R' is the radius of the path's curvature. Even at the top of the hill, the car's velocity and the radius of the hill's curvature determine the necessary centripetal force to maintain it on track, which is balanced by the gravitational pull.
Normal Force
The concept of normal force \(F_n\) is often introduced in physics to describe the force exerted by a surface to support the weight of an object resting on it. This force is always perpendicular to the surface. In the context of our exercise, the normal force comes into play as the car reaches the top of the hill.

Since the centripetal and gravitational forces are acting downwards while the car is at the highest point of its trajectory, the normal force counteracts a part of these forces. By subtracting the centripetal force from the gravitational force, we find a net force. The hill exerts an equal and opposite force to this net force, which is the normal force. The car, similarly, exerts a force on the hill equal in magnitude and opposite in direction to the normal force. In our step-by-step solution, this explains why the force exerted by the car on the hill is \(80.27 \text{ N}\) upwards.

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Most popular questions from this chapter

A centrifuge in a medical laboratory rotates at an angular speed of 3600 rpm (revolutions per minute). When switched off, it rotates 60.0 times before coming to rest. Find the constant angular acceleration of the centrifuge.

In a department store toy display, a small disk (disk 1) of radius \(0.100 \mathrm{~m}\) is driven by a motor and turns a larger disk (disk 2) of radius \(0.500 \mathrm{~m}\). Disk 2 , in turn, drives disk 3 , whose radius is \(1.00 \mathrm{~m}\). The three disks are in contact and there is no slipping. Disk 3 is observed to sweep through one complete revolution every \(30.0 \mathrm{~s}\) a) What is the angular speed of disk \(3 ?\) b) What is the ratio of the tangential velocities of the rims of the three disks? c) What is the angular speed of disks 1 and \(2 ?\) d) If the motor malfunctions, resulting in an angular acceleration of \(0.100 \mathrm{rad} / \mathrm{s}^{2}\) for disk 1 , what are disks 2 and 3's angular accelerations?

Gear \(A\), with a mass of \(1.00 \mathrm{~kg}\) and a radius of \(55.0 \mathrm{~cm}\) is in contact with gear \(\mathrm{B}\), with a mass of \(0.500 \mathrm{~kg}\) and a radius of \(30.0 \mathrm{~cm} .\) The gears do not slip with respect to each other as they rotate. Gear A rotates at 120. rpm and slows to 60.0 rpm in \(3.00 \mathrm{~s}\). How many rotations does gear B undergo during this time interval?

A CD starts from rest and speeds up to the operating angular frequency of the CD player. Compare the angular velocity and acceleration of a point on the edge of the CD to those of a point halfway between the center and the edge of the CD. Do the same for the linear velocity and acceleration.

A monster truck has tires with a diameter of \(1.10 \mathrm{~m}\) and is traveling at \(35.8 \mathrm{~m} / \mathrm{s}\). After the brakes are applied, the truck slows uniformly and is brought to rest after the tires rotate through 40.2 turns. a) What is the initial angular speed of the tires? b) What is the angular acceleration of the tires? c) What distance does the truck travel before coming to rest?
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