The spring constant of the spring is 1040 N/m
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Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.
[tex]\boxed {F = k \times \Delta x}[/tex]
F = Force ( N )
k = Spring Constant ( N/m )
Δx = Extension ( m )
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The formula for finding Young's Modulus is as follows:
[tex]\boxed {E = \frac{F / A}{\Delta x / x_o}}[/tex]
E = Young's Modulus ( N/m² )
F = Force ( N )
A = Cross-Sectional Area ( m² )
Δx = Extension ( m )
x = Initial Length ( m )
Let us now tackle the problem !
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Given:
mass of the object = m = 2.5 kg
initial speed of the object = u = 1.2 m/s
final speed of the object = v = 0 m/s
compression of the spring = x = 5.0 cm = 0.05 m
work done by friction = W_f = -0.50 J
Unknown:
spring constant of the spring = k = ?
Solution:
We will use formula as follows:
[tex]W_f = ( Ek' + Ep' ) - ( Ek + Ep )[/tex]
[tex]W_f = ( 0 + \frac{1}{2}kx^2 ) - ( \frac{1}{2}mv^2 + 0 )[/tex]
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2 - W_f[/tex]
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + W_f[/tex]
[tex]kx^2 = mv^2 + 2W_f[/tex]
[tex]k = ( mv^2 + 2W_f ) \div x^2[/tex]
[tex]k = ( 2.5(1.2)^2 + 2(-0.50) ) \div (0.05)^2[/tex]
[tex]k = ( 2.6 ) \div ( 0.05 )^2[/tex]
[tex]k = 1040 \texttt { N/m}[/tex]
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Grade: College
Subject: Physics
Chapter: Elasticity
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Keywords: Elasticity , Diameter , Concrete , Column , Load , Compressed , Stretched , Modulus , Young
