/*! 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 23 Express the units of the force c... [FREE SOLUTION] | 91Ó°ÊÓ

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Express the units of the force constant of a spring in \(\mathrm{SI}\) base units.

Short Answer

Expert verified
The force constant of a spring in SI base units is expressed as kilograms per second squared (kg/s^2).

Step by step solution

01

Understand the units of the spring constant

Knowing that the spring constant \(k\) is typically measured in Newtons per meter (N/m), we recognize that Newton (N) and meter (m) are both SI units. However, Newton is a derived SI unit (not a base unit), so our task is to further break it down into base units.
02

Break down Newtons into base units

By definition, 1 Newton (N) is the force required to accelerate a one-kilogram mass by one meter per second squared. This gives us the relation 1 N = 1 kg*m/s^2.
03

Express the spring constant in base units

Given the spring constant is represented in N/m, substituting our relation from step 2 allows us to express the spring constant in SI base units: k = N/m = kg*m/s^2*m^-1 = kg/s^2.

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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 is a key concept in physics, particularly within Hooke's Law, which states that the force needed to extend or compress a spring is proportional to the distance it is stretched or compressed. This relationship can be mathematically described as \( F = kx \), where \( F \) represents the force, \( x \) is the displacement of the spring from its natural length, and \( k \) is the spring constant.

The spring constant \( k \) has the units of Newton per meter \((\text{N/m})\). It essentially measures the stiffness of the spring: a higher \( k \) value means a stiffer spring that requires more force to extend or compress. The spring constant can vary for different materials and spring designs. Understanding the spring constant allows us to predict how springs will behave in various forces and environments.

When solving problems related to springs, knowing the spring constant helps in evaluating how much potential energy is stored or how much work is needed to stretch the spring.
Base Units
In the International System of Units (SI), base units are fundamental units from which other units are derived. There are seven SI base units covering different physical quantities, including the meter for length, kilogram for mass, and second for time.

Derived units, like the Newton, are combinations of these base units. It is crucial to understand base units, especially when breaking down more complex measurements into simpler, fundamental parts. For example, the spring constant, expressed as Newton per meter \((\text{N/m})\), ultimately relates to the base units of kilogram (kg), meter (m), and second (s).

This knowledge is particularly useful when performing unit conversions or analyzing formulas and equations in physics. It gives deeper insight into the relationships between different physical phenomena.
Newton
The Newton (N) is the SI unit of force. Named after Sir Isaac Newton, it quantifies the amount of force needed to accelerate a 1-kilogram mass by 1 meter per second squared. Hence, we can express 1 Newton as \( 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2 \).

The Newton is an example of a derived unit, as it originates from the base units for mass (kilogram), length (meter), and time (second). Understanding the breakdown of Newtons into base units is crucial for solving physics problems that involve force and motion. This conversion underlies many scientific calculations, from designing safety features in vehicles to predicting the behavior of celestial bodies.
Force Unit Conversion
Force unit conversion is an important skill in physics. It helps us navigate between different units of measurement and understand their relationships. To convert force units into their base units, we need to break them down according to their definitions. For example, a Newton is broken down into its base units as \( 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2 \).

When converting the spring constant from Newtons per meter \((\text{N/m})\) to SI base units, we use this breakdown: \( k = \frac{\text{N}}{\text{m}} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}} = \text{kg/s}^2 \). This process simplifies understanding and solving problems involving forces, allowing us to sum forces, work with different systems, and understand coupled motion phenomena.

Mastering force unit conversion integrates knowledge of both basic physics concepts and practical problem-solving techniques.

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Most popular questions from this chapter

For saving energy, bicycling and walking are far more efficient means of transportation than is travel by automobile. For example, when riding at \(10.0 \mathrm{mi} / \mathrm{h}\) a cyclist uses food energy at a rate of about \(400 \mathrm{kcal} / \mathrm{h}\) above what he would use if merely sitting still. (In exercise physiology, power is often measured in kcal/h rather than in watts. Here \(1 \mathrm{kcal}=1\) nutritionist's Calorie \(=4186 \mathrm{J} .\) Walking at \(3.00 \mathrm{mi} / \mathrm{h}\) requires about \(220 \mathrm{kcal} / \mathrm{h} .\) It is interesting to compare these values with the energy consumption required for travel by car. Gasoline yields about \(1.30 \times\) \(10^{8} \mathrm{J} /\) gal. Find the fuel economy in equivalent miles per gallon for a person (a) walking, and (b) bicycling.

You can think of the work-kinetic energy theorem as a second theory of motion, parallel to Newton's laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a) and (b) separately from parts \((\mathrm{c})\) and \((\mathrm{d})\) to compare the predictions of the two theories. In a rifle barrel, a \(15.0-\mathrm{g}\) bullet is accelerated from rest to a speed of \(780 \mathrm{m} / \mathrm{s}\). (a) Find the work that is done on the bullet. (b) If the rifle barrel is \(72.0 \mathrm{cm}\) long, find the magnitude of the average total force that acted on it, as \(F=W /(\Delta r \cos \theta) .\) (c) Find the constant acceleration of a bullet that starts from rest and gains a speed of \(780 \mathrm{m} / \mathrm{s}\) over a distance of \(72.0 \mathrm{cm} .\) (d) If the bullet has mass \(15.0 \mathrm{g},\) find the total force that acted on it as \(\Sigma F=m a.\)

A \(15.0\) -kg block is dragged over a rough, horizontal surface by a \(70.0\) - \(\mathrm{N}\) force acting at \(20.0^{\circ}\) above the horizontal. The block is displaced \(5.00 \mathrm{~m}\), and the coefficient of kinetic friction is \(0.300\). Find the work done on the block by (a) the 70 - \(\mathrm{N}\) force, (b) the normal force, and (c) the gravitational force. (d) What is the increase in internal energy of the block- surface system due to friction? (e) Find the total change in the block's kinetic energy.

A 2100 -kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact with the top of the beam, and it drives the beam \(12.0 \mathrm{cm}\) farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

Vector A has a magnitude of 5.00 units, and B has a magnitude of 9.00 units. The two vectors make an angle of \(50.0^{\circ}\) with each other. Find \(\mathbf{A} \cdot \mathbf{B}.\)

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