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2 years ago

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

is found to have stretched 0.0275 m from its original length. The mass is then pushed up 0.0735 m from its equilibrium position and released. What is the kinetic energy of the mass at the instant it passes back through its equilibrium position?

2 answers:

ch4aika [34]2 years ago

8 0

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

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<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

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<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

$\texttt{ }$

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

brilliants [131]2 years ago

5 0

Answer:

KE= 1.20 J

Explanation:

Given that

m = 1.25 kg

x= 0.0275 m

A= 0.0735 m

F= K x = m g

k = mg/x

$k=\dfrac{1.25\times 9.81}{0.0275 }\ N/m$

k=445.9 N/m

This is the spring constant.

Kinetic energy of mass when it passes through equilibrium given as

$KE=\dfrac{kA^2}{2}$

By putting the values

$KE=\dfrac{445.9\times 0.0735^2}{2}$

KE= 1.20 J

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