/*! 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

A block of mass \(m\) is released from rest at the top of an incline that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction between the block and incline is \(\mu_{k}\). The top of the incline is a vertical distance \(h\) above the bottom of the incline. Derive an expression for the work \(W_{f}\) done on the block by friction as it travels from the top of the incline to the bottom. When \(\alpha\) is decreased, does the magnitude of \(W_{f}\) increase or decrease?

Your physics book is resting in front of you on a horizontal table in the campus library. You push the book over to your friend, who is seated at the other side of the table, 0.400 m north and 0.300 m east of you. If you push the book in a straight line to your friend, friction does \(-4.8 \mathrm{~J}\) of work on the book. If instead you push the book \(0.400 \mathrm{~m}\) due north and then \(0.300 \mathrm{~m}\) due east, how much work is done by friction?

Stopping Distance. A car is traveling on a level road with speed \(v_{0}\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of \(v_{0}, g,\) and the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the tires and the road. ( b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

A 4.80 kg watermelon is dropped from rest from the roof of an 18.0 m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what are the watermelon’s (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

A net force along the \(x\) -axis that has \(x\) -component \(F_{x}=-12.0 \mathrm{~N}+\left(0.300 \mathrm{~N} / \mathrm{m}^{2}\right) x^{2}\) is applied to a \(5.00 \mathrm{~kg}\) object that is initially at the origin and moving in the \(-x\) -direction with a speed of \(6.00 \mathrm{~m} / \mathrm{s}\). What is the speed of the object when it reaches the point \(x=5.00 \mathrm{~m} ?\)

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