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Sagot :
Given:
Mass, m = 1.4 kg
Spring constant, k = 140 N/m
x1 = 0.12 m
Let's find the speed of the block when the elongation is x = 0.06 m
Apply the Conservation of Energy formula:
[tex]\frac{1}{2}mv^2=\frac{1}{2}k(x^2_1-x^2_2)[/tex]Since we are to find the speed, rewrite the formula for v:
[tex]v=\sqrt[]{\frac{k(x^2_1-x^2_2)}{m}}[/tex]Where:
k = 140 N/m
x1 = 0.12 m
x2 = 0.06 m
m = 1.4 kg
Hence, we have:
[tex]\begin{gathered} v=\sqrt[]{\frac{140(0.12^2-0.06^2)}{1.4}} \\ \\ v=\sqrt[]{\frac{1.512}{1.4}} \\ \\ v=1.039\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the block when the elongation is x=0.06m is = 1.039 m/s.
ANSWER:
1.039 m/s
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