Question

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

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

Step-by-step solution

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 \(370\, m\).

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, \[F_g = (1000\, kg) \times (9.81\, m/s^2) = 9810\, N\].

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, \[F_c = \dfrac{(1000\, kg) \times (60.0\, m/s)^2}{370\, m} = 9729.73\, N\]

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.

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.

Key concepts

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—represented by the letter 'g'—is 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’s 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.