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7. Friction, Inclines, Systems

Intro to Springs (Hooke's Law)

Physics

7. Friction, Inclines, Systems

Intro to Springs (Hooke's Law)

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5:17 minutes

Problem 59

Textbook Question

A spring of equilibrium length L₁ and spring constant k₁ hangs from the ceiling. Mass m₁ is suspended from its lower end. Then a second spring, with equilibrium length L₂ and spring constant k₂, is hung from the bottom of m₁. Mass m₂ is suspended from this second spring. How far is m₂ below the ceiling?

Verified step by step guidance

Step 1: Understand the system. The problem involves two springs and two masses. The first spring (spring 1) has an equilibrium length L₁ and spring constant k₁, and it supports mass m₁. The second spring (spring 2) has an equilibrium length L₂ and spring constant k₂, and it supports mass m₂. The goal is to find the total distance from the ceiling to mass m₂ when the system is in equilibrium.

Step 2: Analyze the forces acting on mass m₁. At equilibrium, the upward force exerted by spring 1 equals the downward gravitational force on mass m₁ plus the force exerted by spring 2 due to the weight of mass m₂. Mathematically, this can be expressed as:

k_{1} (x 1) = m_{1} g + k_{2} (x 2)

where x₁ is the extension of spring 1 and x₂ is the extension of spring 2.

Step 3: Analyze the forces acting on mass m₂. At equilibrium, the upward force exerted by spring 2 equals the downward gravitational force on mass m₂. This can be expressed as:

k_{2} (x 2) = m_{2} g

Solve this equation to find x₂, the extension of spring 2.

Step 4: Substitute the value of x₂ into the equation from Step 2 to solve for x₁, the extension of spring 1. This will give you the total extension of spring 1 due to the combined forces of m₁ and the force transmitted by spring 2.

Step 5: Calculate the total distance from the ceiling to mass m₂. This is the sum of the equilibrium lengths of both springs (L₁ and L₂) and their respective extensions (x₁ and x₂). Mathematically:

Distance = L_{1} + x 1 + L_{2} + x 2

This will give the total distance of m₂ below the ceiling.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its extension or compression from its equilibrium position, mathematically expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement. This principle is essential for understanding how the springs in the problem will stretch under the weight of the masses attached to them.

Equilibrium Position

The equilibrium position of a spring is the point at which the spring is neither compressed nor extended, meaning the net force acting on it is zero. In the context of the problem, the equilibrium lengths L₁ and L₂ represent the lengths of the springs when no external forces are applied, which is crucial for calculating the total displacement caused by the suspended masses.

Gravitational Force

Gravitational force is the attractive force between two masses, calculated using Newton's law of universal gravitation, F = mg, where m is the mass and g is the acceleration due to gravity. In this scenario, the gravitational forces acting on masses m₁ and m₂ will determine how much each spring stretches, ultimately affecting the position of m₂ relative to the ceiling.

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Related Practice

Textbook Question

A 12 kg weather rocket generates a thrust of 200 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 550 N/m, is anchored to the ground. Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed?

Textbook Question

A 35-cm-long vertical spring has one end fixed on the floor. Placing a 2.2 kg physics textbook on the spring compresses it to a length of 29 cm. What is the spring constant?

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Textbook Question

A 5.0 kg mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in FIGURE EX9.28. The scale reads in newtons. The scale reads 20 N when the lower spring has been compressed by 2.0 cm. What is the value of the spring constant for the lower spring?

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Multiple Choice

A vertical spring is originally 60 cm long. When you attach a 5 kg object to it, the spring stretches to 70 cm.

(a) Find the force constant on the spring.
(b) You now attach an additional 10 kg to the spring. Find its new length.

(For the multiple choice selection, answer only part (b). Use g=10 m/s².)

A 30 g mass is attached to one end of a 10-cm-long spring. The other end of the spring is connected to a frictionless pivot on a frictionless, horizontal surface. Spinning the mass around in a circle at 90 rpm causes the spring to stretch to a length of 12 cm. What is the value of the spring constant?

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Textbook Question

A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm. How long is the spring when a 3.0 kg mass is suspended from it?

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