/*! 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 73 You are asked to design spring b... [FREE SOLUTION] | 91Ó°ÊÓ

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You are asked to design spring bumpers for the walls of a parking garage. A freely rolling \(1200 \mathrm{~kg}\) car moving at \(0.65 \mathrm{~m} / \mathrm{s}\) is to compress the spring no more than \(0.090 \mathrm{~m}\) before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

Short Answer

Expert verified
The spring constant should be approximately \(63,000 \, N/m\).

Step by step solution

01

Understand and note given quantities

Let's first make note of the given quantities. The mass of the car, m, is 1200 kg. The speed of the car, v, is 0.65 m/s. The maximum compression of the spring \(x_{max}\) is 0.090 m.
02

Calculate car's initial kinetic energy

Before hitting the spring, the car has kinetic energy. This is calculated as \(\frac{1}{2} m v^2\). Plugging in the given values gives us \(\frac{1}{2} \times 1200 \, kg \times (0.65 \, m/s)^2 = 253.5 \, J\). This will be the amount of energy stored in the spring at maximum compression.
03

Apply conservation of energy

When the spring is at its maximum compression, the kinetic energy of the car has been fully converted to potential energy of the spring, given as \(\frac{1}{2} k x_{max}^2\). Setting this equal to the initial kinetic energy yields \(\frac{1}{2} k (0.090 \, m)^2 = 253.5 \, J\).
04

Solve for spring constant

Finally, solving for k from the above equation, we get \(k = \frac{2 \times 253.5 \, J}{(0.090 \, m)^2} = 63,000 \, N/m\) (rounded to three significant figures). So, the spring constant should be 63,000 N/m.

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Key Concepts

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

Conservation of Energy
Understanding the conservation of energy principle is crucial when studying physics, and it's particularly important when dealing with moving objects and springs. The principle states that energy cannot be created or destroyed in an isolated system; it can only be transformed from one form to another.

In our spring bumper scenario, when a car hits the spring, this principle tells us that the car's energy hasn't disappeared. Instead, the kinetic energy - the energy due to the car's motion - is converted into potential energy stored in the spring. At maximum compression, all the car's initial kinetic energy is assumed to have converted into the potential energy of the spring, which enables us to calculate the required spring constant.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is quantified by the formula \( K = \frac{1}{2} m v^2 \), where \( m \) is the object's mass and \( v \) is its velocity.

For our problem involving a rolling car, we initially calculate the car's kinetic energy before it contacts the spring. This energy is the resource we use for stopping the car. Kinetic energy is a simple yet powerful concept in physics that directly relates to an object's capacity to do work on another object – in this case, compressing a spring.
Potential Energy
Potential energy, in contrast to kinetic energy, is the energy stored within an object due to its position, arrangement, or state. A parked car on a hill and a drawn bow are both examples of objects with significant potential energy.

In scenarios involving springs, potential energy is present when the spring is either compressed or stretched from its natural length. This stored energy can be harnessed to perform work, which is key when designing mechanisms such as our parking garage spring bumpers.
Spring Potential Energy
Spring potential energy is a specific form of potential energy that is stored when a spring is compressed or stretched from its equilibrium position. The amount of energy stored in a compressed or stretched spring can be described by the formula \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the spring's rest position.

When the car in our example stops, all its kinetic energy is now converted into spring potential energy. By setting the spring's maximum potential energy equal to the car's initial kinetic energy, we can solve for the unknown spring constant, which determines the bumper's effectiveness.
Hooke's Law
Hooke's law is fundamental to understanding how springs behave when forces are applied to them. It states that the force needed to extend or compress a spring by some distance \( x \) scales linearly with respect to that distance, which can be expressed with the formula \( F = -k x \).

The \( k \) in this formula is known as the spring constant, and it measures the stiffness of the spring. A higher \( k \) value means a stiffer spring, and that's precisely what we need to figure out for our parking garage bumpers. Hooke's law, combined with the conservation of energy, allows us to calculate the ideal spring constant to ensure a car can be stopped safely without excessive compression of the spring.

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Most popular questions from this chapter

Using a cable with a tension of \(1350 \mathrm{~N}\), a tow truck pulls a car \(5.00 \mathrm{~km}\) along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

A \(4.00 \mathrm{~kg}\) block of ice is placed against one end of a horizontal spring that is fixed at the other end, has force constant \(k=200 \mathrm{~N} / \mathrm{m}\) and is compressed \(0.025 \mathrm{~m}\). The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

Varying Coefficient of Friction. A box is sliding with a speed of \(4.50 \mathrm{~m} / \mathrm{s}\) on a horizontal surface when, at point \(P,\) it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at \(12.5 \mathrm{~m}\) past point \(P .\) (a) Use the work-energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of \(0.100 ?\)

A baseball has a mass of 0.145 kg. (a) In batting practice a batter hits a ball that is sitting at rest on top of a post. The ball leaves the post with a horizontal speed of \(30.0 \mathrm{~m} / \mathrm{s}\). How much work did the force applied by the bat do on the ball? (b) During a game the same batter swings at a ball thrown by the pitcher and hits a line drive. Just before the ball is hit it is traveling at a speed of \(20.0 \mathrm{~m} / \mathrm{s},\) and just after it is hit it is traveling in the opposite direction at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\). Whatis the total work done on the baseball by the force exerted by the bat? (c) How do the results of parts (a) and (b) compare? Explain.

Meteor Crater. About 50,000 years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about \(1.4 \times 10^{8} \mathrm{~kg}\) (around 150,000 tons) and hit the ground at a speed of \(12 \mathrm{~km} / \mathrm{s}\). (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0 megaton nuclear bomb? (A megaton bomb releases the same amount of energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{~J}\) of energy.)

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