Ch 08: Dynamics II: Motion in a Plane

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 08: Dynamics II: Motion in a Plane

Problem 29

Chapter 8, Problem 29

You are driving your 1800 kg car at 25 m/s over a circular hill that has a radius of 150 m. A deer running across the road causes you to hit the brakes hard while right at the summit of the hill, and you start to skid. The coefficient of kinetic friction between your tires and the road is 0.75. What is the magnitude of your acceleration as you begin to slow?

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Identify the forces acting on the car at the summit of the hill. These include the gravitational force \( F_g = m g \), the normal force \( F_N \), and the frictional force \( F_f \). The frictional force is given by \( F_f = \mu_k F_N \), where \( \mu_k \) is the coefficient of kinetic friction.

At the summit of the hill, the car is in circular motion. The net force acting towards the center of the circular path is the centripetal force \( F_c \), which is provided by the difference between the gravitational force and the normal force: \( F_c = F_g - F_N \). Use \( F_c = \frac{m v^2}{r} \) to relate the centripetal force to the car's mass \( m \), velocity \( v \), and the radius of the hill \( r \).

Solve for the normal force \( F_N \) using the equation \( F_N = F_g - F_c \). Substitute \( F_g = m g \) and \( F_c = \frac{m v^2}{r} \) into this equation to find \( F_N \).

Determine the frictional force \( F_f \) using \( F_f = \mu_k F_N \). This frictional force is responsible for the car's deceleration as it begins to skid.

Use Newton's second law \( F = m a \) to find the acceleration \( a \). The net force causing the deceleration is the frictional force \( F_f \), so \( a = \frac{F_f}{m} \). Substitute \( F_f = \mu_k F_N \) and the expression for \( F_N \) to calculate the magnitude of the acceleration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Acceleration

Centripetal acceleration is the acceleration directed towards the center of a circular path that an object follows. It is necessary for maintaining circular motion and is calculated using the formula a_c = v^2 / r, where v is the velocity and r is the radius of the circular path. In this scenario, the car experiences centripetal acceleration as it travels over the hill.

Friction and Kinetic Friction

Friction is the force that opposes the relative motion of two surfaces in contact. Kinetic friction specifically refers to the frictional force acting on moving objects, which can be calculated using the formula F_k = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force. In this case, the coefficient of kinetic friction between the tires and the road will determine how quickly the car can decelerate.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, expressed as F = m * a. This principle is crucial for calculating the car's deceleration as it skids, as the net force will be the difference between the frictional force and the gravitational force acting on the car.

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