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
[tex]k=\dfrac{1.25\times 9.81}{0.0275 }\ N/m[/tex]
k=445.9 N/m
This is the spring constant.
Kinetic energy of mass when it passes through equilibrium given as
[tex]KE=\dfrac{kA^2}{2}[/tex]
By putting the values
[tex]KE=\dfrac{445.9\times 0.0735^2}{2}[/tex]
KE= 1.20 J
The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J
[tex]\texttt{ }[/tex]
Let's recall Elastic Potential Energy formula as follows:
[tex]\boxed{E_p = \frac{1}{2}k x^2}[/tex]
where:
Ep = elastic potential energy ( J )
k = spring constant ( N/m )
x = spring extension ( compression ) ( m )
Let us now tackle the problem!
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Given:
mass of object = m = 1.25 kg
initial extension = x = 0.0275 m
final extension = x' = 0.0735 - 0.0275 = 0.0460 m
Asked:
kinetic energy = Ek = ?
Solution:
Firstly , we will calculate the spring constant by using Hooke's Law as follows:
[tex]F = k x[/tex]
[tex]mg = k x[/tex]
[tex]k = mg \div x[/tex]
[tex]k = 1.25(9.8) \div 0.0275[/tex]
[tex]k = 445 \frac{5}{11} \texttt{ N/m}[/tex]
[tex]\texttt{ }[/tex]
Next , we will use Conservation of Energy formula to solve this problem:
[tex]Ep_1 + Ek_1 = Ep_2 + Ek_2[/tex]
[tex]\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek[/tex]
[tex]Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2[/tex]
[tex]Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh[/tex]
[tex]Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)[/tex]
[tex]\boxed {Ek \approx 1.20 \texttt{ J}}[/tex]
[tex]\texttt{ }[/tex]
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Grade: High School
Subject: Physics
Chapter: Elasticity
