/*! 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 58 A truck with mass \(m\) has a br... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A truck with mass \(m\) has a brake failure while going down an icy mountain road of constant downward slope angle \(\alpha\) (Fig. \(\overline{P 7} .58\) ). 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 \(\beta .\) The truck ramp has a soft sand surface for which the coefficient of rolling friction is \(\mu_{\mathrm{r}}\). What is the distance that the truck moves up the ramp before coming to a halt? Solve by energy methods.

Short Answer

Expert verified
The distance that the truck moves up the ramp before coming to a halt is given by \(d = \frac{0.5v_{0}^{2} + gL\sin(\alpha)}{g(\sin(\beta) + \mu_{\mathrm{r}})}\)

Step by step solution

01

Set up the energy equation

By conservation of energy, the initial energy of the truck must equal to its final energy. The initial energy of the truck consists of gravitational potential energy \( mgh_1 \) where \( h_1 = L \sin(\alpha) \) is the height from the bottom of the uphill ramp to the spot where the truck starts to move downhill, and kinetic energy \( 0.5 \cdot m \cdot v_{0}^{2} \). The final energy of the truck is when it is at rest, its potential energy \( mgh_2 \) where \( h_2 \) is the final height and energy due to rolling friction \( \mu_{\mathrm{r}}mgd \) where \( d \) is the distance the truck moves up the ramp. So the energy equation is written as \( 0.5 \cdot m \cdot v_{0}^{2} + mgh_1 = mgh_2 + \mu_{\mathrm{r}}mgd \)
02

Substituting for the heights

The heights are determined by the geometry of the problem. We replace \( h_1 \) with \( L \sin(\alpha) \) and \( h_2 \) with \( d \sin(\beta) \). So the equation becomes \( 0.5 \cdot m \cdot v_{0}^{2} + mgL\sin(\alpha) = mgd\sin(\beta) + \mu_{\mathrm{r}}mgd \)
03

Solving for distance \(d\)

We can simplify the equation by dividing through by \(mg\), then collect all terms involving \(d\) to one side. This gives us \(d = \frac{0.5v_{0}^{2} + gL\sin(\alpha)}{g\sin(\beta) + \mu_{\mathrm{r}}g}\).
04

Simplify the final equation

We can factor out \(g\) from the denominator to simplify the equation to \( d = \frac{0.5v_{0}^{2} + gL\sin(\alpha)}{g(\sin(\beta) + \mu_{\mathrm{r}})} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic and Potential Energy
In physics, energy can appear in various forms, including kinetic and potential energy.

Kinetic energy is the energy possessed by an object due to its motion. It's calculated with the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the moving object and \( v \) is its velocity. As an object’s speed increases, so does its kinetic energy.

On the flip side, potential energy is the energy stored in an object due to its position or configuration. Gravitational potential energy in particular relates to the position of an object in a gravitational field and is given by \( PE = mgh \), where \( h \) is the height above a reference level, and \( g \) is the acceleration due to gravity.

In the context of the truck problem, as the truck moves downhill, it converts potential energy to kinetic energy. Conversely, when the truck ascends the ramp, its kinetic energy is converted back into potential energy until it comes to a stop.
Rolling Friction
Rolling friction, also known as rolling resistance, is the resistive force that occurs when an object rolls over a surface. It is much weaker than sliding friction that you would have between two surfaces rubbing against each other.

The coefficient of rolling friction, denoted as \( \mu_{r} \), characterizes how much friction there is. It combines factors like the texture of the surface, the rigidity of the wheel, and the deformation that occurs where they meet.

In physics exercises, rolling friction plays a critical role when calculating the deceleration of objects, such as wheels on a vehicle. In our truck problem, the question specifically asks us to account for rolling friction on the sand surface of the runaway truck ramp where it directly impacts how far the truck travels before halting.
Mechanical Energy Conservation
Mechanical energy conservation is a principle in physics stating that the total mechanical energy of an isolated system remains constant if only conservative forces are acting. Conservative forces, like gravity, don't change the total mechanical energy - they just convert it from one form to another, as is the case with potential and kinetic energy.

However, when non-conservative forces, like friction, are part of the equation, they dissipate mechanical energy as heat or other forms of non-mechanical energy. In our truck scenario, energy is not conserved because rolling friction is doing work.

To solve for the distance the truck moves up the ramp before coming to a halt, we consider the mechanical energy at the start and at the end, with the work done by friction included in the energy balance.
Physics Problem-Solving
Solving physics problems often involves breaking down complex situations into manageable pieces, identifying relevant concepts, and applying equations that express physical laws. A widely adopted approach is the energy method, which leans on the principle of conservation of energy.

In this method, we set up an energy balance equation which expresses that the total energy at one point in time (or place) must equal the total energy at another, accounting for the work done by any non-conservative forces.

For the truck problem, we applied this approach by equating the truck's energy down the hill with its energy as it ascends the ramp. The presence of rolling friction requires us to include an additional term for the work done by this force, showing how a proper understanding of physics concepts plays a key role in problem-solving.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a 65 kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that \(100 \%\) of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.

If a fish is attached to a vertical spring and slowly lowered to its equilibrium position, it is found to stretch the spring by an amount \(d\). If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance \(d\) and the mass \(m\) of the fish.)

During the calibration process, the cantilever is observed to deflect by \(0.10 \mathrm{nm}\) when a force of \(3.0 \mathrm{pN}\) is applied to it. What deflection of the cantilever would correspond to a force of \(6.0 \mathrm{pN} ?\) (a) \(0.07 \mathrm{nm}\) (b) \(0.14 \mathrm{nm} ;\) (c) \(0.20 \mathrm{nm} ;\) (d) \(0.40 \mathrm{nm}\).

\(\mathrm{CP}\) A \(0.300 \mathrm{~kg}\) potato is tied to a string with length \(2.50 \mathrm{~m}\) and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

A small block of mass \(m\) on a horizontal frictionless surface is attached to a horizontal spring that has force constant \(k .\) The block is pushed against the spring, compressing the spring a distance \(d\). The block is released, and it moves back and forth on the end of the spring. During its motion, what is the maximum speed of the block?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.