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A person who weighs 670 \(\mathrm{N}\) steps onto a spring scale in the bathroom, and the spring compresses by 0.79 \(\mathrm{cm} .\) (a) What is the spring constant? \((\text { b } \text { bhat is the weight of another person who compresses the }\) spring by 0.34 \(\mathrm{cm} ?\)

Short Answer

Expert verified
(a) 84810.13 N/m (b) 288.35 N.

Step by step solution

01

Understand Hooke's Law

To find the spring constant, apply Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement or compression of the spring. The formula is given by: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement of the spring.
02

Convert Displacement to Meters

The displacement given is 0.79 cm. We need to convert this to meters since SI units are used in the formula. Therefore, \( x = 0.79 \text{ cm} = 0.0079 \text{ m} \).
03

Calculate Spring Constant

Given that the weight provides the force, \( F = 670 \, \mathrm{N} \). We can rearrange the formula \( F = kx \) to solve for the spring constant: \( k = \frac{F}{x} \). Substituting the values, \( k = \frac{670}{0.0079} \approx 84810.13 \, \mathrm{N/m} \).
04

Calculate New Compression Force

For the second person, we know the compression is 0.34 cm. Convert this to meters: \( x = 0.34 \text{ cm} = 0.0034 \text{ m} \). Use the spring constant found: \( k = 84810.13 \, \mathrm{N/m} \). The weight of the new person is \( F = kx \).
05

Solve for New Weight

Substitute the values into the formula: \( F = 84810.13 \cdot 0.0034 \approx 288.35 \, \mathrm{N} \). This is the weight of the other person.

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Key Concepts

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

Spring Constant
The spring constant, often symbolized as \( k \), is a measure of a spring’s stiffness. In the language of physics, it's how much force is needed to move the spring a specific distance. Hooke's Law is the cornerstone concept here, expressed as \( F = kx\), where \( F \) is the force applied, and \( x \) is the spring's displacement.

This equation tells us that the force exerted by a spring is directly proportional to its displacement. The spring constant \( k \) is essentially the proportionality factor.

Here’s a simple way to interpret it:
* **A larger spring constant** means the spring is stiffer and harder to compress.
* **A smaller spring constant** means the spring is looser and easier to compress.

For example, using Hooke's Law, if you measure the force applied to a spring and the spring's displacement, you can rearrange the formula to solve for \( k \): \( k = \frac{F}{x} \). Understanding this principle is essential for solving many physics problems involving springs.
Force and Displacement
In mechanics, force and displacement are closely related concepts. Displacement, denoted as \( x \) in Hooke's Law, is the measure of how much a spring is compressed or extended from its resting position.

Force, noted as \( F \), is commonly measured in newtons and represents the weight or mass exerted on an object. In terms of springs, it’s the force that causes the spring to either compress or stretch.

When applying Hooke’s Law, understanding these two concepts is crucial:
* **When a force is applied**, the spring either compresses or stretches, depending on the direction.
* **The amount it compresses or stretches**, known as displacement, directly relates to the force applied through the spring constant.

The displacement must always be measured from the spring's equilibrium position, the point at which it is neither compressed nor stretched, to accurately apply Hooke’s Law.
Unit Conversion in Physics
Unit conversion is a vital skill in physics, crucial for ensuring that all quantities used in calculations are consistent with each other.

For example, in Hooke's Law problems, displacement is typically measured in meters (). If you’re given displacement in centimeters, like 0.79 cm in the original exercise, you must convert this to meters before plugging it into \( F = kx \).

Here’s a quick guide on converting units:
* **To convert from centimeters to meters**, divide by 100 because there are 100 cm in a meter. Hence, 0.79 cm becomes 0.0079 m.
* **Always check your units** before performing calculations to avoid errors.

Understanding and performing unit conversions accurately ensures calculations conform with standard International System of Units (SI units), which is vital for consistency and accuracy in physics.

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Most popular questions from this chapter

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a ram spring \((k=32000 \mathrm{N} / \mathrm{m})\) . The mass of this spring may be ignored The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\)when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a sling-shot fires a stone. Setting up for this stunt involves stretching the bands by 3.00 \(\mathrm{m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of 26.8 \(\mathrm{m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

A rifle fires a \(2.10 \times 10^{-2}-\mathrm{kg}\) pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{m}\) from its unstrained length. The pellet rises to a maximum height of 6.10 \(\mathrm{m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

\(\mathrm{A} 1.00 \times 10^{-2}-\mathrm{kg}\) block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 124 \(\mathrm{N} / \mathrm{m}\) . The block is shoved parallel to the spring axis and is given an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

An 86.0 -kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{N} / \mathrm{m}\) . He accidentally slips and falls freely for 0.750 \(\mathrm{m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

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