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A spring scale hung from the ceiling stretches by \(6.4 \mathrm{cm}\) when a \(1.0 \mathrm{kg}\) mass is hung from it. The \(1.0 \mathrm{kg}\) mass is removed and replaced with a \(1.5 \mathrm{kg}\) mass. What is the stretch of the spring?

Short Answer

Expert verified
The stretch of the spring when a 1.5 kg mass is hung from it is 0.096 m or 9.6 cm.

Step by step solution

01

Determine the Spring Constant (k)

Firstly, Hooke's law equation is: \(F = kx\), where \(F\) is the force applied, \(k\) is the spring constant and \(x\) is the displacement or stretch of the spring. Here, the force exerted by the 1.0 kg mass can be calculated using the gravitational force equation \(F = mg\), where \(m\) is the mass and \(g\) is acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\)). So, \(F = 1.0 \mathrm{kg} * 9.8 \mathrm{m/s^2} = 9.8 \mathrm{N}\). Also, the given displacement is 6.4 cm, which needs to be converted into meters by dividing by 100, so \(x = 6.4 \mathrm{cm} / 100 = 0.064 \mathrm{m}\). Therefore, by rearranging the Hooke's law equation to find \(k\), we obtain \(k = F/x = 9.8 \mathrm{N} / 0.064 \mathrm{m} = 153.125 \mathrm{N/m}\).
02

Calculate the Stretch for the 1.5 kg Mass

Now, determine the stretch of the spring when the 1.5 kg mass is hung. The force exerted by the 1.5 kg mass is \(F = 1.5 \mathrm{kg} * 9.8 \mathrm{m/s^2} = 14.7 \mathrm{N}\). Using the Hooke's law equation again and substituting \(F = 14.7 \mathrm{N}\) and \(k = 153.125 \mathrm{N/m}\), the stretch \(x\) of the spring can be found by \(x = F/k = 14.7 \mathrm{N} / 153.125 \mathrm{N/m} = 0.096 \mathrm{m}\).

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Key Concepts

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

Spring Constant
Understanding the spring constant is crucial when analyzing the elastic behavior of a spring under the application of a force. The spring constant, denoted as 'k', is a measure of the stiffness of a spring. It tells us the amount of force needed to stretch or compress the spring by a certain distance. Higher spring constant values indicate a stiffer spring, requiring more force to produce the same amount of displacement as a spring with a lower spring constant.

Let's consider a practical scenario to illustrate this concept. Imagine you're holding two springs, one stiff like the one used in car suspensions, and the other loose, like that of a small toy. If you apply the same force to both, you'll notice a smaller change in length in the stiffer spring due to its higher spring constant. In Hooke's Law, expressed as the equation \( F = kx \), 'k' embodies this property. The equation shows us that the force (F) exerted by or on the spring is directly proportional to the displacement (x) experienced by the spring, with 'k' being the constant of proportionality.

In our exercise, solving for 'k' using the given force (due to the weight of the mass) and displacement helped us find the spring constant. Once we have 'k', we can then predict how the spring will behave under different forces, which is fundamental for designing systems that depend on precise spring responses.
Gravitational Force
Gravitational force is a natural phenomenon by which all things with mass or energy are attracted toward one another. On Earth, this force causes objects to have weight. It is computed using the formula \( F = mg \), where 'F' represents the gravitational force, 'm' the mass of the object, and 'g' the acceleration due to gravity, which on Earth's surface is approximately \( 9.8 \mathrm{m/s^2} \).

This concept becomes very practical when working with spring scales. The weight of an object due to gravitational force causes the spring scale to stretch. The more massive the object, the greater the force it exerts on the spring, leading to a greater displacement. In educational settings or laboratory experiments, gravity's role in physics problems like the one from our exercise allows students to see the direct relationship between an object's mass, the force it exerts due to gravity, and the resulting motion or deformation (such as the stretching of a spring).

By calculating the gravitational force for the two different masses in our exercise, it becomes apparent how the change in mass affects the stretch of the spring—a key principle for understanding weight measurement and the functioning of spring-driven mechanisms.
Displacement
Displacement in the context of springs and Hooke's Law refers to the change in length of the spring from its original, unstressed length. This can be either a stretch (elongation) or a compression (shortening). It's important to note that displacement is a vector quantity—it has both magnitude and direction. In the case of a vertical spring, the direction is typically downward for stretching and upward for compression, relative to the spring's equilibrium position.

When an object is suspended from a spring scale, the scale's length changes; this change is the displacement. In our textbook problem, when we hang masses from a spring, we observe this as a stretching of the spring scale. Displacement is directly related to the force applied to the spring: the more significant the force (in this case, from a heavier mass), the more the spring stretches, resulting in more considerable displacement. This relationship allows us to use displacement as a key indicator of the forces at play.

It is important for students to convert the displacement measurement to the appropriate units to align with other parameters in their calculations, as we did in our exercise by converting centimeters to meters. Displacement not only aids us in understanding forces and spring behavior but also in applications across physics, engineering, and even everyday life, such as measuring the weight of produce with a spring scale at the market.

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Most popular questions from this chapter

On your first trip to Planet \(X\) you happen to take along a 200 g mass, a 40.0-cm-long spring, a meter stick, and a stopwatch. You're curious about the free-fall acceleration on Planet \(X,\) where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by \(31.2 \mathrm{cm} .\) You then pull the mass down \(10.0 \mathrm{cm}\) and release it. With the stopwatch you find that 10 oscillations take \(14.5 \mathrm{s}\). Can you now satisfy your curiosity?

The New England Merchants Bank Building in Boston is \(152 \mathrm{m}\) high. On windy days it sways with a frequency of \(0.17 \mathrm{Hz},\) and the acceleration of the top of the building can reach \(2.0 \%\) of the free-fall acceleration, enough to cause discomfort for occupants. What is the total distance, side to side, that the top of the building moves during such an oscillation?

The motion of a nightingale's wingtips can be modeled as simple harmonic motion. In one study, the tips of a bird's wings were found to move up and down with an amplitude of \(8.8 \mathrm{cm}\) and a period of 0.82 s. What are the wingtips' (a) maximum speed and (b) maximum acceleration?

Calculate and draw an accurate displacement graph from \(t=0 \mathrm{s}\) to \(t=10 \mathrm{s}\) of a damped oscillator having a frequency of \(1.0 \mathrm{Hz}\) and a time constant of \(4.0 \mathrm{s}\)

A spring with spring constant \(15.0 \mathrm{N} / \mathrm{m}\) hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest. It is then pulled down \(6.00 \mathrm{cm}\) and released. If the ball makes 30 oscillations in 20.0 s, what are its (a) mass and (b) maximum speed?

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