/*! 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 41 $$ \begin{array}{l}{\text { (I... [FREE SOLUTION] | 91Ó°ÊÓ

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$$ \begin{array}{l}{\text { (II) Uphill escape ramps are sometimes provided to the }} \\ {\text { side of steep downhill highways for trucks with overheated }} \\ {\text { brakes. For a simple } 11^{\circ} \text { upward ramp, what length would be }} \\ {\text { needed for a runaway truck traveling } 140 \mathrm{km} / \mathrm{h} \text { ? Note the }}\end{array} $$ $$ \begin{array}{l}{\text { large size of your calculated length. (If sand is used for the bed }} \\ {\text { of the ramp, its length can be reduced by a factor of about } 2 .}\end{array} $$

Short Answer

Expert verified
The uphill ramp should be approximately 423.8 meters long.

Step by step solution

01

Convert Speed to Meters Per Second

First, we need to convert the speed from kilometers per hour to meters per second. The given speed is 140 km/h. To convert it, use the conversion factor: \[ \frac{140 \text{ km/h} \times 1000 \text{ m/km}}{3600 \text{ s/h}} = \frac{140,000}{3600} \text{ m/s} \approx 38.89 \text{ m/s}. \] Thus, the truck's speed is approximately 38.89 m/s.
02

Calculate Kinetic Energy

Next, calculate the truck's kinetic energy using the formula \( KE = \frac{1}{2}mv^2 \). Since the mass \( m \) is not given, we will assume it cancels out later. Therefore, \[ KE = \frac{1}{2}m(38.89 \text{ m/s})^2. \]
03

Calculate Potential Energy at the Ramp's Elevation

At the top of the ramp, the truck will have converted all of its kinetic energy into gravitational potential energy, which is given by \( PE = mgh \), where \( h \) is the height of the ramp. Equating the kinetic energy to the potential energy gives \[ \frac{1}{2}m(38.89)^2 = mgh. \]
04

Relate Height to Ramp Length

Use the trigonometric relationship \( \sin 11^\circ = \frac{h}{L} \) to express height in terms of length: \[ h = L \sin 11^\circ. \] Substitute this into the potential energy equation to get \[ \frac{1}{2}(38.89)^2 = gL \sin 11^\circ. \]
05

Solve for the Ramp Length

Solving for \( L \), we substitute in \( g \approx 9.81 \text{ m/s}^2 \) and \( \sin 11^\circ \approx 0.1908 \): \[ \frac{1}{2}(38.89)^2 = 9.81 \times L \times 0.1908. \] Then, \[ L = \frac{\frac{1}{2}(38.89)^2}{9.81 \times 0.1908}. \] Calculating this gives \[ L \approx 423.8 \text{ meters}. \]

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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 has due to its motion. Whenever an object moves, it possesses kinetic energy, which depends on its mass and speed. To calculate kinetic energy, we use the formula:\[ KE = \frac{1}{2}mv^2 \]Here, \( m \) stands for mass, and \( v \) is the velocity of the object. This formula shows that kinetic energy increases with the square of the velocity, meaning that small increases in speed can lead to significant increases in kinetic energy.
For example, in the problem, even without knowing the truck's mass, we calculate the kinetic energy knowing it travels at 38.89 m/s.
Understanding kinetic energy is crucial because it's involved every time an object accelerates, stops, or collides with another object, forming a central concept of mechanics.
Potential Energy
Potential energy is the stored energy of position and is particularly significant when dealing with heights. For gravitational potential energy, which is the common type we encounter, the formula is:\[ PE = mgh \]Where \( m \) is the mass, \( g \) the acceleration due to gravity (approximately 9.81 \( m/s^2 \)), and \( h \) the height above a reference point. In the uphill ramp problem, as the truck ascends the ramp, it converts all of its kinetic energy into potential energy.
This transformation occurs because energy is conserved, meaning total energy remains constant even when it shifts forms. Hence, understanding potential energy helps predict how far up the ramp the truck can travel before stopping.
Trigonometry
Trigonometry, the study of triangles, is a handy tool for solving physics problems involving angles, forces, and ramp heights. In the ramp problem, we apply it to relate the ramp's length and height using the sine function:\[ \sin(\theta) = \frac{h}{L} \]Here, \( \theta \) is the ramp angle (11°), \( h \) the height, and \( L \) the hypotenuse, or the ramp's length. By knowing the angle and wanting to find the required length, trigonometry allows us to plug into this relationship. With \( \sin(11°) \approx 0.1908 \), it's crucial for calculating the ramp's actual length as potential energy depends on height. Thus, mastering trigonometry enriches problem-solving strategies in mechanics.
Mechanics
Mechanics is the branch of physics dealing with motion and forces. It encompasses concepts like kinetic and potential energy, momentum, and the laws of motion. In the context of our ramp problem, mechanics explains the truck's movement and the energy changes as it ascends the ramp.
By combining principles of potential and kinetic energy with gravity and ramp angles, mechanics enables us to predict and influence how objects interact with their environment. Applications range from designing safety features like escape ramps to planning efficient transport routes. Hence, a solid grasp of mechanics empowers learners to understand and solve practical real-world problems using physics.
Physics Education
Physics education aims to impart a solid understanding of fundamental concepts in physics, helping students apply these to various real-world situations. Whether it's understanding energy transformations or utilizing mathematics in physics, education equips learners with skills to tackle complex problems systematically.
In exercises like calculating the length of a ramp, physics education highlights step-by-step problem-solving techniques. These include converting units, applying conservation of energy, and using trigonometric relationships, all while promoting critical thinking and analytical skills.
  • Emphasizes core physics principles
  • Focuses on real-life applications
  • Encourages systematic problem solving
Ultimately, physics education fosters a deep comprehension of the universe while preparing students for diverse scientific challenges.

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Most popular questions from this chapter

A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 g's. Calculate the force on a \(65-\mathrm{kg}\) person accelerating at this rate. What distance is traveled if brought to rest at this rate from \(95 \mathrm{~km} / \mathrm{h} ?\)

(II) Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.8 \(\mathrm{m}\) and is released with a speed of 13 \(\mathrm{m} / \mathrm{s} .\)

A super high-speed 14-car Italian train has a mass of 640 metric tons \((640,000 \mathrm{~kg})\). It can exert a maximum force of \(400 \mathrm{kN}\) horizontally against the tracks, whereas at maximum constant velocity \((300 \mathrm{~km} / \mathrm{h})\), it exerts a force of about \(150 \mathrm{kN}\). Calculate \((a)\) its maximum acceleration, and \((b)\) estimate the force of friction and air resistance at top speed.

In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to push grocery carts up the ramps and it is obviously desirable that this not be too difficult. The engineer has done a survey and found that almost no one complains if the force required is no more than \(18 \mathrm{~N}\). Ignoring friction, at what maximum angle \(\theta\) should the ramps be built, assuming a full \(25-\mathrm{kg}\) grocery cart?

(III) A heavy steel cable of length \(\ell\) and mass \(M\) passes over a small massless, frictionless pulley. (a) If a length \(y\) hangs on one side of the pulley (so \(\ell-y\) hangs on the other side), calculate the acceleration of the cable as a function of \(y .\) (b) Assuming the cable starts from rest with length \(y_{0}\) on one side of the pulley, determine the velocity \(v_{\mathrm{r}}\) at the moment the whole cable has fallen from the pulley. (c) Evaluate \(v_{\mathrm{f}}\) for \(y_{0}=\frac{2}{3} \ell\) . [Hint: Use the chain rule, \(d v / d t=(d v / d y)(d y / d t),\) and integrate. \(]\)

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