/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 A truck with mass \(m\) has a br... [FREE SOLUTION] | 91Ó°ÊÓ

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A truck with mass \(m\) has a brake failure while going down an ?cy mountain road of constant downward slope angle \(\alpha\) (Fig. 7.40\()\) Initially the truck is moving downhill at speed \(v_{0}\) - After careening downhill a distance \(L\) with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle \(\boldsymbol{\beta}\) . The truck rump has a soft sand suffice for which the coefficient of rolling friction is \(\mu_{r}\) What is the distance that the truck moves up the rump before coming to a halt? Solve using energy methods.

Short Answer

Expert verified
The truck moves up the ramp for a distance \( x = \frac{v_0^2}{2g( \sin \beta + \mu_r \cos \beta )} \).

Step by step solution

01

Initial Energy

Calculate the initial kinetic energy of the truck as it starts moving downhill. The kinetic energy (KE) is given by the formula:\[ KE_{initial} = \frac{1}{2}mv_0^2 \]
02

Work Done by Friction

As the truck moves up the ramp, it does work against the frictional force. The frictional force can be calculated as:\[ f_{friction} = \mu_r mg \cos \beta \]The work done by this force over distance \(x\) is:\[ W_{friction} = f_{friction} \cdot x = \mu_r mg \cos \beta \cdot x \]
03

Change in Gravitational Potential Energy

As the truck moves up the ramp to a height \(h\), its gravitational potential energy will increase by:\[ \Delta PE = mgh \]Since \(h\) can be expressed in terms of \(x\) (distance moved up the ramp) and \(\beta\) (ramp angle) as \(h = x \sin \beta \), we substitute:\[ \Delta PE = mgx \sin \beta \]
04

Energy Conservation Equation

Apply the principle of conservation of energy, which states the initial kinetic energy is equal to the sum of the work done by friction and the change in potential energy:\[ \frac{1}{2}mv_0^2 = mgx \sin \beta + \mu_r mg \cos \beta \cdot x \]
05

Solve for Distance x

Factor out \(x\) from the right side of the equation:\[ \frac{1}{2}mv_0^2 = mgx \left( \sin \beta + \mu_r \cos \beta \right) \]Solve for \(x\):\[ x = \frac{v_0^2}{2g( \sin \beta + \mu_r \cos \beta )} \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on two main factors: the mass of the object and its velocity. The formula to calculate kinetic energy is given by:
  • \( KE = \frac{1}{2} mv^2 \)
where \( KE \) represents kinetic energy, \( m \) is the mass, and \( v \) is the velocity.
In the context of the truck scenario, the initial kinetic energy is pivotal in understanding how long the truck will move uphill before stopping. It's calculated using the truck's mass and its speed as it begins its downward journey.
Kinetic energy is often exchanged with other forms of energy, such as potential energy or can be lost as work done against friction, indicating its vital role in energy conservation scenarios.
Gravitational Potential Energy
Gravitational potential energy is energy stored in an object due to its position in a gravitational field. It is directly proportional to the height of the object relative to a reference point, usually ground level. The formula used to calculate potential energy is:
  • \( PE = mgh \)
where \( PE \) stands for potential energy, \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
For the truck scenario, as it moves up the ramp, its height increases, converting some kinetic energy into gravitational potential energy. The change in potential energy is \( \Delta PE = mgx \sin \beta \), illustrating how the height achieved is dependent on the distance moved and the slope angle, \( \beta \).
This concept is vital for comprehending how energy transformations occur in systems.
Frictional Force
A frictional force is a force exerted by a surface as an object moves across it. This force opposes the motion and acts in the opposite direction to the movement of the object, often resulting in energy loss from the system. The frictional force can be modeled as:
  • \( f_{friction} = \mu_r mg \cos \beta \)
where \( \mu_r \) is the coefficient of rolling friction, \( m \) is the mass of the object, and \( \beta \) is the angle of the slope.
In the case of the truck, as it travels up the ramp, it experiences a force opposing its movement due to the soft sand, represented by the coefficient \( \mu_r \).
The work done against this frictional force, \( W_{friction} = \mu_r mg \cos \beta \cdot x \), plays a crucial role in the total energy considerations, as it contributes to the stopping distance calculated for the truck.
Energy Methods
Energy methods offer a systematic way to analyze mechanical systems by considering the different energy forms and how they convert into each other.
The conservation of energy principle is foundational in these methods, stating that in a closed system, energy cannot be created or destroyed; it only changes forms.
In the truck exercise, energy methods help us relate the initial kinetic energy of the truck to both the work done by friction and the change in gravitational potential energy along the ramp.
  • Initial kinetic energy is being converted into potential energy and work done against friction.
  • The sum of the energy forms at different points must equal initially available energy.
This approach simplifies complex motion problems, providing a clear path to determining variables like distance traveled before stopping, as shown by setting up the final equation for the truck's motion:\[ \frac{1}{2} mv_0^2 = mgx \sin \beta + \mu_r mg \cos \beta \cdot x \]
These methods are crucial for efficiently solving energy transformation problems.

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Most popular questions from this chapter

Gravity in One Dimension. Two point masses, \(m_{1}\) and \(m_{2}\) , lie on the \(x\) -axis, with \(m_{1}\) held in place at the origin and \(m_{2}\) at position \(x\) and free to move. The gravitational potential energy of these masses is found to be \(U(x)=-G m_{1} m_{2} / x,\) where \(G\) is a constant (called the gravitational constant). You'll learn more about gravitation in Chapter 12 . Find the \(x\) -component of the force acting on \(m_{2}\) due to \(m_{1} .\) Is this force attractive or repulsive? How do you know?

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