Answer:
0.012 J
Explanation:
Spring constant K = 250 N/m given
Initial compression that is [tex]X_i=5\ mm\ =0.005\ m[/tex]
Additional compression that = 6 mm =0.006 m
So final compression that is [tex]X_f =0.005+0.006=0.0.011m[/tex]
Work done is given by change in potential energy of spring
So [tex]w=\Delta PE=\frac{1}{2}\times K\times (X_f^2-X_i^2)[/tex]
So work done [tex]w=\frac{1}{2}\times 250\times (0.011^2-0.005^2)=0.012 \ J[/tex]
0.012 J of work is done by the spring force to ready the pen for writing
[tex]\texttt{ }[/tex]
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 )
[tex]\texttt{ }[/tex]
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 !
[tex]\texttt{ }[/tex]
Given:
spring constant = k = 250 N/m
initial compression = x₀ = 5.0 mm = 5.0 × 10⁻³ m
final compression = x = 5.0 + 6.0 = 11.0 mm = 11.0 × 10⁻³ m
Asked:
work = W = ?
Solution:
[tex]W = \Delta Ep[/tex]
[tex]W = Ep - Ep_o[/tex]
[tex]W = \frac{1}{2}kx^2 - \frac{1}{2}kx_o^2[/tex]
[tex]W = \frac{1}{2}k( x^2 - x_o^2 )[/tex]
[tex]W = \frac{1}{2}(250)[ (11.0 \times 10^{-3})^2 - (5.0 \times 10^{-3})^2][/tex]
[tex]W \approx 0.012 \texttt{ J}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Elasticity
