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What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 7.5 cm?

Short Answer

Expert verified
The spring constant is approximately 8889 N/m.

Step by step solution

01

Understand the problem

We are given the amount of elastic potential energy stored in a spring and the displacement by which the spring is compressed. Our task is to find the spring constant, often denoted as \( k \).
02

Write down the formula for elastic potential energy

The formula for the elastic potential energy \( E \) stored in a spring is given by: \[ E = \frac{1}{2} k x^2 \]where \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.
03

Identify the known variables

From the problem, we know:- \( E = 25 \text{ J} \) (elastic potential energy)- \( x = 7.5 \text{ cm} = 0.075 \text{ m} \) (displacement, converted from cm to m)
04

Substitute known values into the formula

Substitute the known values into the formula for elastic potential energy:\[25 = \frac{1}{2} k (0.075)^2\]
05

Solve for the spring constant \( k \)

First, simplify the equation:\[25 = \frac{1}{2} k (0.005625)\]Multiply through by 2 to eliminate the fraction:\[50 = k \, \times \, 0.005625\]Finally, solve for \( k \) by dividing both sides by 0.005625:\[k = \frac{50}{0.005625} \approx 8888.89 \]So, \( k \approx 8889 \text{ N/m}\) after rounding.

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials, like springs, due to their deformation. When you compress or stretch a spring, you store energy in it. This stored energy can later do work, for example, by moving an object. Elastic potential energy enables mechanisms ranging from toy springs to complex machinery to function. One key formula associated with elastic potential energy is:
  • \( E = \frac{1}{2} k x^2 \)
    • The variable \( E \) represents the elastic potential energy in joules (J). The spring constant, \( k \), measures how stiff the spring is, while \( x \) indicates how much the spring is stretched or compressed from its rest position, measured in meters.
      Understanding elastic potential energy is crucial when examining how springs store energy and the factors affecting it such as the spring's stiffness and the amount of deformation.
Spring Compression
Spring compression refers to the act of shortening a spring from its equilibrium state by an external force. This physical change results in storing elastic potential energy within the spring. The degree of compression affects how much energy can be stored and depends on the spring's characteristics. Imagine compressing a spring with your hand. The more you compress:
  • The harder it becomes to push further.
  • The more energy the spring stores.
Spring compression can be quantified using displacement, \( x \), which is the distance the spring is compressed from its normal, unstressed position.
Converting units is essential when calculating problems involving spring compression, ensuring consistent measurements, such as converting centimeters to meters for standard calculations.
Hooke's Law
Hooke's Law is foundational in understanding spring dynamics and describes how the force needed to extend or compress a spring is proportional to the distance compressed or stretched. According to Hooke's Law:
  • The force exerted by the spring, \( F \), is given by: \( F = -kx \)
The negative sign indicates that the force exerted by the spring is in the opposite direction of the applied force. The spring constant \( k \), determines how stiff the spring is. A higher \( k \) means a stiffer spring, requiring more force for the same displacement.
This relation explains why springs behave predictably under tension or compression and helps in solving problems by linking force, displacement, and the spring's properties. Understanding Hooke's Law aids in manipulating and utilizing springs in practical ways, from basic applications in doors to advanced uses in engineering systems.

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Most popular questions from this chapter

You push a \(2.0 \mathrm{~kg}\) block against a horizontal spring, compressing the spring by \(15 \mathrm{~cm}\). Then you release the block, and the spring sends it sliding across a tabletop. It stops \(75 \mathrm{~cm}\) from where you released it. The spring constant is \(200 \mathrm{~N} / \mathrm{m}\). What is the block-table coefficient of kinetic friction?

A \(70 \mathrm{~kg}\) firefighter slides, from rest, \(4.3 \mathrm{~m}\) down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides, so that the average frictional force of the pole on her is \(500 \mathrm{~N}\) upward, what is her speed just before reaching the ground floor?

A \(3.2 \mathrm{~kg}\) sloth hangs \(3.0 \mathrm{~m}\) above the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point \(y=0\) to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

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