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In the Back to the Future movies (https:I/openstaxcollege.org/l/21bactofutclip) DeLorean car of mass \(1230 \mathrm{kg}\) travels at 88 miles per hour to venture back to the future. (a) What is the kinetic energy of the DeLorean? (b) What spring constant would be needed to stop this DeLorean in a distance of \(0.1 \mathrm{m}\) ?

Short Answer

Expert verified
The kinetic energy of the DeLorean is \( K.E. \) Joules. The required spring constant is \( k \) Newtons per meter.

Step by step solution

01

Convert Velocity to Meters per Second

The given velocity is 88 mph. We know that 1 mile equals 1609.34 meters and 1 hour equals 3600 seconds. Using these conversions, the velocity in meters per second can be found as follows: \( v = 88 \times \frac{1609.34 \, \mathrm{m}}{3600 \, \mathrm{sec}} \)
02

Calculate Kinetic Energy

The formula for kinetic energy is \( K.E. = \frac{1}{2} m v^2 \), where m is the mass and v is the velocity. The mass of the DeLorean car is given as 1230 kg. Using these values, one can find the kinetic energy.
03

Calculate Spring Constant

The formula for spring constant is \( k = \frac{m v^2}{x^2} \) where m is the mass, v is the velocity and x is the distance. Here, the distance is given as 0.1 m. Using these values, calculate the spring constant.

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Key Concepts

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

Physics Problem Solving
Physics problem solving is all about understanding the concepts, applying the formulas correctly, and often involves converting units to match the standard international system (SI units). It's a critical skill in physics to be able to break down a problem into smaller parts and solve it step by step.

In our example involving the famous DeLorean from 'Back to the Future', the first step is always to clarify the quantities we have and what units they are in. The car's speed is initially given in miles per hour, which we need to convert to meters per second because SI units make calculations straightforward and allow for the direct application of physics formulas.

In tackling the problem, a systematic approach is adopted. The conversion of units laid the groundwork for the subsequent calculations, specifically the determination of kinetic energy and the required spring constant. This adherence to a consistent method ensures accuracy and helps students to better understand and solve physics problems.
Energy Conversion
Energy conversion in physics refers to the process of changing one form of energy into another. In the context of the DeLorean problem, we're looking at converting kinetic energy into potential energy stored in a spring. Kinetic energy is the energy of movement, which depends on the mass and velocity of the moving object.

When the moving DeLorean is brought to a stop, its kinetic energy is not lost but is instead transferred into the spring, compressing it and storing this energy as elastic potential energy. The relationship between kinetic energy and potential energy in a spring is a prime example of the conservation of energy, which is a fundamental concept in physics stating that energy cannot be created or destroyed, only converted from one form into another.

A deeper understanding of energy conversion can significantly enhance a student's ability to analyze real-world physics problems, where multiple energy transformations often take place.
Spring Constant Formula
The spring constant formula is an essential aspect of Hooke's law, which defines the force required to compress or extend a spring by a certain amount. The spring constant, denoted by 'k', is a measure of the stiffness of the spring. The formula, which plays a crucial role in our DeLorean example, is given by:
\[ k = \frac{F}{x} \]
where 'F' is the force exerted on the spring and 'x' is the displacement from its equilibrium position. However, in the case of stopping the car, we use a modified version of the formula:
\[ k = \frac{m v^2}{x^2} \]
Here, we ascertain the spring constant 'k' using the mass of the vehicle 'm', the square of its velocity 'v', and the distance 'x' by which the spring is compressed. A high spring constant indicates a stiff spring, which, in this scenario, means a greater ability to absorb the car's kinetic energy within a small compression distance. Understanding the spring constant allows us to predict how a spring will behave under various forces, an important concept in engineering and physics alike.

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Most popular questions from this chapter

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back \(50 \mathrm{cm}\) and holds it in position with a force of 150 N. If the mass of the arrow is \(50 \mathrm{g}\) and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

In a coyote/Road Runner cartoon clip (https://openstaxcollege.org/I/21coyroadcarcl) , a spring expands quickly and sends the coyote into a rock. If the spring extended \(5 \mathrm{m}\) and sent the coyote of mass 20 kg to a speed of \(15 \mathrm{m} / \mathrm{s},\) (a) what is the spring constant of this spring? (b) If the coyote were sent vertically into the air with the energy given to him by the spring, how high could he go if there were no non-conservative forces?

Find the force corresponding to the potential energy \(U(x)=-a / x+b / x^{2}\)

A particle of mass \(2.0 \mathrm{kg}\) moves under the influence of the force \(F(x)=(3 / \sqrt{x}) \mathrm{N} .\) If its speed at \(x=2.0 \mathrm{m}\) is \(v=6.0 \mathrm{m} / \mathrm{s},\) what is its speed at \(x=7.0 \mathrm{m} ?\)

A projectile of mass 2 kg is fired with a speed of 20 \(\mathrm{m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with respect to the horizontal. (a) Calculate the initial total energy of the projectile given that the reference point of zero gravitational potential energy at the launch position. (b) Calculate the kinetic energy at the highest vertical position of the projectile. (c) Calculate the gravitational potential energy at the highest vertical position. (d) Calculate the maximum height that the projectile reaches. Compare this result by solving the same problem using your knowledge of projectile motion.

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