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Chapter 10: Problem 5

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? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{cm} ?\)

Short Answer

Expert verified
(a) The spring constant is approximately 84810.13 N/m. (b) The weight of the second person is approximately 288.35 N.

Step by step solution

01

Understanding Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring, which is described by the formula: \[ F = kx \]where \( F \) is the force applied (in Newtons), \( k \) is the spring constant (in N/m), and \( x \) is the displacement (in meters).
02

Convert Units

Before substituting into the formula, it's essential to convert all units to match. Here, convert the displacement from centimeters to meters:\[ 0.79\, \text{cm} = 0.0079\, \text{m} \]
03

Solve for Spring Constant (k)

Using the values obtained, substitute them into Hooke's Law equation to find \( k \):\[ 670\, \text{N} = k \times 0.0079\, \text{m} \]Solve for \( k \):\[ k = \frac{670}{0.0079} \approx 84810.13\, \text{N/m} \]
04

Convert Displacement for Second Person

Convert the displacement for the second person from centimeters to meters:\[ 0.34\, \text{cm} = 0.0034\, \text{m} \]
05

Calculate Force for Second Person

Using the spring constant found in Step 3, calculate the force (weight) for the second person using Hooke's Law:\[ F = kx = 84810.13 \times 0.0034 \]\[ F \approx 288.35\, \text{N} \]

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Key Concepts

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

Spring Constant
When dealing with springs, the spring constant is a fundamental concept. It tells us how stiff or stretchy a spring is. According to Hooke's Law, the force needed to compress or extend a spring is proportional to the distance it is stretched or compressed.
  • The formula for Hooke's Law is: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement.
  • The spring constant \( k \) is measured in Newtons per meter (N/m).
  • A larger \( k \) value means a stiffer spring that requires more force to compress or extend.

In our problem, the spring constant is calculated using the force exerted by a person's weight and the displacement of the spring. This calculation helps us determine how easily a spring is compressed under a given force. Understanding this lets us predict how a spring will behave under different forces.
Force and Displacement
Force and displacement in the context of springs are closely related. The displacement is the amount the spring is compressed or extended from its rest position.
  • In our example, displacement must be converted to meters because the unit for the spring constant requires this.
  • The force exerted by the spring is equal to the applied weight when a person stands on a spring scale.
  • Hooke's Law simplifies the relationship: \( F = kx \)
By knowing the displacement and the spring constant, one can easily find the force exerted by the spring. In the case of another person stepping on the spring, understanding this relationship allows us to calculate their weight if the spring compresses by a known amount.
Unit Conversion
Unit conversion is an essential step when applying Hooke's Law. Often, measurements need to be in compatible units for calculations to make sense.
  • In this exercise, displacement was initially given in centimeters.
  • To use the Hooke's Law formula, it was necessary to convert this to meters.
  • The conversion from centimeters to meters is done by dividing by 100, since 1 meter = 100 centimeters.

Consistent unit conversion ensures that calculations are accurate and correspond to real-world measurements. This step is crucial, especially in physics problems, where unit discrepancies can lead to incorrect results.

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Most popular questions from this chapter

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m},\) released from rest, and subsequently oscillate back and forth with an angular frequency of \(11.3 \mathrm{rad} / \mathrm{s} .\) What is the speed of the object at the instant when the spring is stretched by \(0.048 \mathrm{m}\) relative to its unstrained length?

You are given the task of opening an antiquated "light lock," which is unlocked by shining red light pulses of a certain frequency for a long duration of time into a light sensor on the lock. You are given a red laser pointer, a spring of unstretched length \(L=15.0 \mathrm{cm}\) and spring constant \(k=7.20 \mathrm{N} / \mathrm{m},\) a sheet of steel \(\left(\rho=7.60 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 0.125 inches thick, and some tools. You come up with the idea to take a piece of the steel sheet (of mass \(m\) ), cut a slot in it, and hang it from the spring. If you shine the laser through the slot and onto the sensor, and then stretch the spring and let it go, the steel plate will oscillate and cause the beam to pass through the slot periodically. (a) Assuming the beam is passing through the slot (and onto the lock's sensor) when the spring-mass system is in equilibrium, how is the frequency at which the light pulses hit the sensor related to the frequency of the spring/mass (i.e., steel plate) system. (b) Based on your answer for (a), what should be the frequency of the spring/mass system if the unlocking frequency is \(3.50 \mathrm{Hz} ?\) (c) What should be the mass \(m\) of the steel plate? (d) Calculate some reasonable dimensions for the steel plate (i.e., they should be consistent with the mass that is required for the spring-mass system). You may assume the material removed to make the slot in the steel plate is of negligible mass.

A spring stretches by \(0.018 \mathrm{m}\) when a \(2.8-\mathrm{kg}\) object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m}\) ) is mounted on the floor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

In preparation for shooting a ball in a pinball machine, a spring \((k=675 \mathrm{N} / \mathrm{m})\) is compressed by \(0.0650 \mathrm{m}\) relative to its unstrained length. The ball \((m=0.0585 \mathrm{kg})\) is at rest against the spring at point \(\mathrm{A} .\) When the spring is released, the ball slides (without rolling). It leaves the spring and arrives at point \(\mathrm{B},\) which is \(0.300 \mathrm{m}\) higher than point \(\mathrm{A} .\) Ignore friction, and find the ball's speed at point \(\mathrm{B}\).
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