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Chapter 13: Problem 8

A \(2000-\mathrm{kg}\) automobile starts from rest at point \(A\) on a \(6^{\circ}\) incline and coasts through a distance of \(150 \mathrm{m}\) to point \(B\). The brakes are then applied, causing the automobile to come to a stop at point \(C\), which is \(20 \mathrm{m}\) from \(B\). Knowing that slipping is impending during the braking period and neglecting air resistance and rolling resistance, determine \((a)\) the speed of the automobile at point \(B,(b)\) the coefficient of static friction between the tires and the road.

Short Answer

Expert verified
(a) The speed at B is approximately 15.17 m/s. (b) The static friction coefficient is approximately 0.164.

Step by step solution

01

Identify the forces and energy changes

Consider the energy conversion happening from point A to B. Since the car starts from rest, its initial kinetic energy at A is zero. As it coasts down the incline, gravitational potential energy is converted into kinetic energy. We can neglect air resistance and rolling resistance.
02

Calculate gravitational potential energy decrease

The car travels a distance of 150 m down a 6掳 incline. Calculate the change in height (h) using \(h = 150 \, \sin(6^\circ)\). Then find the decrease in gravitational potential energy (\(\Delta PE\)) using \(\Delta PE = mgh\), where \(m = 2000\, \mathrm{kg}\) and \(g = 9.81\, \mathrm{m/s^2}\).
03

Relate potential energy to kinetic energy at B

At point B, all the potential energy has been converted to kinetic energy \(KE = \frac{1}{2}mv^2\). Set the decrease in gravitational potential energy equal to the kinetic energy at point B: \(mgh = \frac{1}{2}mv^2\). Solve for the speed \(v\) at point B by canceling \(m\) and rearranging the equation.
04

Apply energy principles to find speed at B

Substitute \(h = 150 \, \sin(6^\circ)\), \(g = 9.81\, \text{m/s}^2\), into the equation and solve for \(v\). The calculation leads to: \(v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times (150 \times \sin(6^\circ))}\).
05

Analyze forces during braking to find friction coefficient

During braking from B to C, use the equation of motion \(v^2 = u^2 + 2a s\) where \(u\) is the initial speed, \(v = 0\) is the final speed, \(a\) is the acceleration (deceleration), and \(s = 20\, \mathrm{m}\) is the distance. Solve for \(a\).
06

Use Newton's second law for deceleration

The stopping force is due to static friction, which equals \(ma = \mu mg \cos(6^\circ)\). Here, \(\mu\) is the coefficient of friction to find, \(\mu mg \cos(6^\circ) = -ma\). Rearrange to solve for \(\mu\).
07

Calculate and verify the friction coefficient

Substitute \(a\) obtained from step 5, \(m = 2000 \, \mathrm{kg}, g = 9.81 \, \mathrm{m/s^2}\) into the friction equation to determine \(\mu\). \(\mu = -\frac{a}{g \cos(6^\circ)}\). Ensure consistency by checking if \(\mu > 0\).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. When an object is at a height above the ground, it has the potential to do work as it falls under the influence of gravity.
In the given exercise, the automobile starts from rest at the top of a 6掳 incline. As it descends the hill, the potential energy is converted into kinetic energy, which increases the car's speed.
To calculate the change in gravitational potential energy as the car travels down the 150 m incline, we use the formula:
  • Height change, \( h = 150\sin(6^\circ) \)
  • Gravitational potential energy change, \( \Delta PE = mgh \)
Where \( m \) is mass (2000 kg), \( g \) is the acceleration due to gravity (9.81 m/s虏), and \( h \) is the height. It's important to understand that as the car loses height, this potential energy transforms into kinetic energy鈥攁n example of energy conservation in action.
Coefficient of Static Friction
The coefficient of static friction (\( \mu_s \)) is a dimensionless value that represents the frictional force resistance when two surfaces are at rest relative to each other. It determines the force needed to initiate movement between the surfaces.
In the context of this exercise, the coefficient of static friction is crucial during the braking phase from point B to C. It's quantified to ensure that the car comes to a stop under controlled frictional forces before sliding occurs.
To find the coefficient of static friction during braking, we must solve the equation derived from Newton's second law:
  • \( ma = \mu mg \cos(6^\circ) \)
This equation relates the deceleration force due to braking to the maximum static friction force. It's important to note that this scenario assumes that the point just before any sliding or slipping occurs, marking the limit of static friction. In this case, solving for \( \mu \) involves knowing the deceleration (\( a \)) found in previous steps and using it to determine the point of impending slip.
Inclined Plane Mechanics
Inclined plane mechanics deal with the analysis of objects moving on a slope, where gravity, normal force, and friction interact. When considering motion on an inclined plane, it's important to resolve forces correctly and understand how they contribute to movement.
The forces at play are often broken down into components parallel and perpendicular to the slope. For an automobile moving on an incline:
  • Gravitational Force: The component of weight pulling the car down the slope, calculated as \( mg \sin(\theta) \).
  • Normal Force: Perpendicular to the surface, calculated as \( mg \cos(\theta) \).
  • Frictional Force: Acts opposite to the direction of motion, involved in the braking phase.
When analyzing the transition of energy from gravitational potential energy to kinetic energy on an inclined plane, knowing the angle of the incline and the forces present is essential. This helps predict how the object will behave as it moves along the slope, crucial for solving problems involving motion dynamics.

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Most popular questions from this chapter

In anticipation of a long \(6^{\circ}\) upgrade, a bus driver accelerates at a constant rate from \(80 \mathrm{km} / \mathrm{h}\) to \(100 \mathrm{km} / \mathrm{h}\) in 8 s while still on a level section of the highway. Knowing that the speed of the bus is \(100 \mathrm{km} / \mathrm{h}\) as it begins to climb the grade at time \(t=0\) and that the driver does not change the setting of the throttle or shift gears, determine (a) the speed of the bus when \(t=10 \mathrm{s},(b)\) the time when the speed is \(60 \mathrm{km} / \mathrm{h}\).

Two identical billiard balls can move freely on a horizontal table. Ball \(A\) has a velocity \(\mathbf{v}_{0}\) as shown and hits ball \(B,\) which is at rest, at a point \(C\) defined by \(\theta=45^{\circ} .\) Knowing that the coefficient of restitution between the two balls is \(e=0.8\) and assuming no friction, determine the velocity of each ball after impact.

A 0.25 -lb ball thrown with a horizontal velocity \(\mathrm{v}_{0}\) strikes a 1.5 -lb plate attached to a vertical wall at a height of \(36 \mathrm{in.}\) above the ground. It is observed that after rebounding, the ball hits the ground at a distance of \(24 \mathrm{in.}\) from the wall when the plate is rigidly attached to the wall (Fig. 1) and at a distance of 10 in. when a foam- rubber mat is placed between the plate and the wall (Fig. 2 ). Determine (a) the coefficient of restitution \(e\) between the ball and the plate, (b) the initial velocity \(v_{0}\) of the ball.

Prove that a force \(F(x, y, z)\) is conservative if, and only if, the following relations are satisfied: $$\frac{\partial F_{x}}{\partial y}=\frac{\partial F_{y}}{\partial x} \quad \frac{\partial F_{y}}{\partial z}=\frac{\partial F_{z}}{\partial y} \quad \frac{\partial F_{z}}{\partial x}=\frac{\partial F_{x}}{\partial z}$$

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop. A 10 -ton truck enters a \(15^{\circ}\) ramp at a high speed \(v_{0}=108 \mathrm{ft} / \mathrm{s}\) and travels for \(6 \mathrm{s}\) before its speed is reduced to \(36 \mathrm{ft} / \mathrm{s}\). Assuming constant deceleration, determine \((a)\) the magnitude of the braking force, \((b)\) the additional time required for the truck to stop. Neglect air resistance and rolling resistance.
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