91Ó°ÊÓ

You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B\), which was stopped at a red light along a road headed down a hill (Fig. 6-74). You find that the slope of the hill is \(\theta=12.0^{\circ}\), that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{1}=18.0 \mathrm{~m} / \mathrm{s}\). With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) \(0.60\) (dry road surface) and (b) \(0.10\) (road surface covered with wet leaves)?

Step by step solution

01

Identify the given values

Slope of the hill, \(\theta = 12.0^{\circ}\), distance between cars, \( d = 24.0 \mathrm{~m}\), initial speed of car A, \( v_1 = 18.0 \mathrm{~m/s}\), coefficient of kinetic friction for part (a), \( \mu_k = 0.60\), and for part (b), \( \mu_k = 0.10\).

02

Determine the forces acting on the car

Calculate the parallel component of gravity: \( F_{g,\parallel} = mg \sin(\theta) \), the normal force: \( N = mg \cos(\theta) \), and the frictional force: \( F_f = \mu_k N \).

03

Calculate net acceleration down the hill

The net force along the incline is \( F_{net} = mg \sin(\theta) - \mu_k mg \cos(\theta)\). The net acceleration is given by \( a = g( \sin(\theta) - \mu_k \cos(\theta))\).

04

Apply kinematic equation to find final speed

Using \[ v_2^2 = v_1^2 + 2 a d \], calculate the final speed \( v_2 \) just before car A hits car B: \[ v_2 = \sqrt{v_1^2 + 2 d g (\sin (\theta) - \mu_k \cos (\theta))}} \].

05

Solve for final speed in case (a)

Substitute \( \mu_k = 0.60 \): \[ v_2 = \sqrt{(18.0 \mathrm{~m/s})^2 + 2 \times 24.0 \mathrm{~m} \times 9.8 \mathrm{~m/s^2} (\sin(12.0^{\circ}) - 0.60 \cos(12.0^{\circ}))}} \].

06

Solve for final speed in case (b)

Substitute \( \mu_k = 0.10 \): \[ v_2 = \sqrt{(18.0 \mathrm{~m/s})^2 + 2 \times 24.0 \mathrm{~m} \times 9.8 \mathrm{~m/s^2} (\sin(12.0^{\circ}) - 0.10 \cos(12.0^{\circ}))}} \].

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Key Concepts

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

Kinematic Equations

Kinematic equations are used to describe the motion of objects. These equations consider the initial velocity, final velocity, acceleration, time, and displacement. In the context of the exercise, we use the kinematic equation \[ v_2^2 = v_1^2 + 2ad \]. This equation helps us find the final speed of car A (\(v_2\)) just before it hits car B. Here,

• \(v_1\) is the initial velocity
• \(a\) is the net acceleration down the hill
• \(d\) is the distance between the cars

By understanding how each variable influences the equation, you can compute the final speed of an object under uniform acceleration.

Frictional Forces

Frictional forces play a crucial role in this problem. Friction is the resistance that one surface or object encounters when moving over another. There are different types of friction, but we are focused on kinetic friction here. Kinetic friction is the force between moving surfaces.

The frictional force \(F_f\) can be calculated using \[ F_f = \mu_k N \]. Where

• \(\mu_k\) is the coefficient of kinetic friction
• \(N\) is the normal force

In our problem, the surface conditions change, so \(\mu_k\) is 0.60 for dry conditions and 0.10 for wet conditions. This directly affects the frictional force and, therefore, the net acceleration down the hill.

Motion on Inclined Planes

Inclined planes are surfaces tilted at an angle relative to the horizontal. When dealing with objects on inclined planes, we must account for the gravitational force components. The forces can be broken down into two components:

• The parallel component along the plane: \(F_{g,\parallel} = mg \sin(\theta)\)
• The perpendicular component normal to the plane: \(F_{g,\perp} = mg \cos(\theta)\)

In our problem, car A is moving down a hill with a slope of 12°. By understanding how gravity affects the motion on an inclined plane, you can determine the net force acting on the car and thus its net acceleration.

Newton's Second Law

Newton's second law describes how the velocity of an object changes when it is subjected to an external force. Newton's second law can be written as \[ F_{net} = ma \], where

• \(F_{net}\) is the net force
• \(m\) is the mass of the object
• \(a\) is the acceleration

In our exercise, the net force (\(F_{net}\)) is the difference between the parallel component of gravity (\(F_{g,\parallel}\)) and the frictional force (\(F_f\)). Therefore, \[ F_{net} = mg \sin(\theta) - \mu_k mg \cos(\theta) \]. This helps us determine the net acceleration acting down the hill as \[ a = g(\sin(\theta) - \mu_k \cos(\theta)) \]. Understanding Newton’s second law allows us to link the forces acting on an object to its acceleration, providing a comprehensive picture of the object's motion.