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Chapter 10: Problem 15

Runaway Truck In Fig. \(10-33\), a runaway truck with failed brakes is moving downgrade at \(130 \mathrm{~km} / \mathrm{h}\) just before the driver steers the truck up a frictionless emergency escape ramp with an in- clination of \(15^{\circ} .\) The truck's mass is \(5000 \mathrm{~kg} .\) (a) What minimum length \(L\) must the ramp have if the truck is to stop (momentarily) along it? (Assume the truck is a particle, and justify that assumption.) Does the minimum length \(L\) increase, decrease, or remain the same if (b) the truck's mass is decreased and (c) its speed is decreased?

Short Answer

Expert verified
(a) 248.2 m. (b) Mass decrease: Length remains the same. (c) Speed decrease: Length decreases.

Step by step solution

01

- Convert Speed to Meters per Second

First, convert the speed from kilometers per hour to meters per second. The truck's speed is given as 130 km/h.Use the conversion: 1 km/h = 1000 m / 3600 sSo,\[130 \text{ km/h} = 130 \times \frac{1000}{3600} \text{ m/s} = 36.11 \text{ m/s} \]
02

- Determine the Kinetic Energy

Calculate the truck's initial kinetic energy. The formula for kinetic energy is:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the truck's mass and \( v \) is its velocity.\[ KE = \frac{1}{2} \times 5000 \text{ kg} \times (36.11 \text{ m/s})^2 \]\[ KE = \frac{1}{2} \times 5000 \times 1304.3 \]\[ KE = 3260750 \text{ J} \]
03

- Relate Kinetic Energy to Gravitational Potential Energy

As the truck moves up the ramp, its kinetic energy converts to gravitational potential energy (GPE). The formula for GPE is:\[ PE = mgh \]where \( h \) is the height gained, \( g \) is the acceleration due to gravity (9.8 m/s虏), and \( m \) is the mass of the truck.
04

- Determine the Height Gained

Since all kinetic energy converts to potential energy at the point where the truck stops momentarily:\[ mgh = KE \]So,\[ 5000 \times 9.8 \times h = 3260750 \]\[ h = \frac{3260750}{5000 \times 9.8} \]\[ h = 66.55 \text{ m} \]
05

- Use Trigonometry to Find Length of Ramp

Using the inclination angle and the height, determine the length of the ramp. The height \( h \) and the length \( L \) form a right triangle with the angle of inclination \( \theta = 15^{\text{o}} \).\[ \tan(\theta) = \frac{h}{L} \]So,\[ L = \frac{h}{\tan(15^{\text{o}})} \]\[ L = \frac{66.55}{\tan(15^{\text{o}})} \]\[ L = \frac{66.55}{0.2679} \]\[ L \thickapprox 248.2 \text{ m} \]
06

- Analyze Effect of Mass Decrease on Length (Part b)

If the truck's mass is decreased, the kinetic energy is proportionally smaller, but since the potential energy also depends on mass, the mass terms cancel out in the energy equations:\[ mgh = \frac{1}{2} mv^2 \]which simplifies to:\[ gh = \frac{1}{2} v^2 \]Therefore, the minimum length \( L \) of the ramp is not affected by a change in mass.
07

- Analyze Effect of Speed Decrease on Length (Part c)

If the truck's speed is decreased, the kinetic energy is proportional to the square of the speed. Lower speed results in lower kinetic energy and thus a lower height gained:\[ h \thickapprox \frac{v^2}{2g} \]Since length \( L \thickapprox \frac{h}{\tan(\theta)} \), reducing speed reduces height \( h \) and thus the length \( L \) of the ramp.

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Key Concepts

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

Kinetic Energy
In physics, kinetic energy (\text{KE}) refers to the energy an object possesses due to its motion. It depends on two factors: mass (\text{m}) and velocity (\text{v}). The mathematical formula to calculate kinetic energy is \(KE = \frac{1}{2} mv^2\). This means that if the mass or velocity of an object increases, the kinetic energy increases as well. For example, in the runaway truck problem, the truck with a mass of 5000 kg moving at a velocity of 36.11 m/s has significant kinetic energy. When calculating, this results in 3260750 Joules (J).
Understanding how to convert units, like converting speed from km/h to m/s, is crucial. This conversion helps ensure that the values are in the correct units for using the kinetic energy formula.
Gravitational Potential Energy
Gravitational potential energy (\text{GPE}) is the energy an object possesses because of its position in a gravitational field. For our purposes, it鈥檚 the energy an object has when it鈥檚 elevated above the ground. The formula for GPE is \(PE = mgh\), where \text{m} is mass, \text{g} is the acceleration due to gravity (approximately 9.8 m/s虏 on Earth), and \text{h} is the height.
When the truck in our problem ascends the ramp, its kinetic energy transforms into gravitational potential energy. This transition means that, as the truck slows down, its height increases until it comes to a temporary stop. This is why calculating the height (\text{h}) is essential in understanding how far up the ramp the truck will travel.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed; it can only change forms. In the truck problem, this means that the truck's kinetic energy will transform completely into gravitational potential energy as it ascends the ramp.
This principle is illustrated by the equation \(mgh = KE\). Here, the kinetic energy (\text{KE}) before the truck starts climbing the ramp translates into gravitational potential energy (\text{GPE}) at its highest point. Because these energies are equal at the point where the truck briefly stops, we use this equality to solve for the height the truck climbs, and consequently, the length of the ramp.
Trigonometry in Physics
Trigonometry plays a crucial role in solving physical problems involving angles and lengths. In our runaway truck scenario, we use trigonometry to determine the minimum length of a ramp with a known inclination angle. The ramp forms a right triangle where:\( \tan(胃) = \frac{h}{L} \). Here, \(胃\) is the angle of inclination (15掳), \(h\) is the height gained, and \(L\) is the length of the ramp.
By rearranging the formula to solve for \(L\), we get \( L = \frac{h}{\tan(\theta)} \). This equation shows how the height of the ramp and the incline angle are related to find the ramp's length, demonstrating how understanding trigonometry is essential for solving real-world physics problems.

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Most popular questions from this chapter

Ballistic Pendulum A bullet of mass \(10 \mathrm{~g}\) strikes a ballistic pendulum of mass \(2.0 \mathrm{~kg}\). The center of mass of the pendulum rises a vertical distance of \(12 \mathrm{~cm}\). Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Block Drawn by Rope A \(3.57 \mathrm{~kg}\) block is drawn at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor by a rope. The force on the block from the rope has a magnitude of \(7.68 \mathrm{~N}\) and is directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

Snowball A \(1.50 \mathrm{~kg}\) snowball is fired from a cliff \(12.5 \mathrm{~m}\) high with an initial velocity of \(14.0 \mathrm{~m} / \mathrm{s}\), directed \(41.0^{\circ}\) above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

Billiard Ball A billiard ball moving at a speed of \(2.2 \mathrm{~m} / \mathrm{s}\) strikes an identical stationary ball a glancing blow. After the collision, one ball is found to be moving at a speed of \(1.1 \mathrm{~m} / \mathrm{s}\) in a direction making a \(60^{\circ}\) angle with the original line of motion. (a) Find the velocity of the other ball. (b) Can the collision be inelastic, given these data?

Block on Incline Collides with Spring In Fig. \(10-37\), a \(12 \mathrm{~kg}\) block is released from rest on a \(30^{\circ}\) frictionless incline. Below the block is a spring that can be compressed \(2.0 \mathrm{~cm}\) by a force of \(270 \mathrm{~N}\). The block momentarily stops when it compresses the spring by \(5.5 \mathrm{~cm} .\) (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches spring?
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