The speed of the block at the displacement from the equilibrium position is 1.062 m/s.
The given parameters:
The speed of the block is calculated by applying the principle of conservation of mechanical energy as shown below;
[tex]\frac{1}{2} mv^2 = \frac{1}{2}kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{800 \times 0.13^2}{12} } \\\\v = 1.062 \ m/s[/tex]
Thus, the speed of the block at the displacement from the equilibrium position is 1.062 m/s.
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Answer:
The speed of the block at the displacement from the equilibrium position is 1.1266 m/s.
Step-by-step explanation:
Solution :
Using principle of conservation of mechanical energy formula to find the speed of the block :
[tex]\begin{gathered} \longrightarrow{\pmb{\sf{\frac{1}{2} mv^2 = \frac{1}{2}kx^2}}}\end{gathered}[/tex]
As per given data information in the question we have :
Substituting all the given values in the formula to find the speed of the block
[tex]\longrightarrow{\sf{ \: \:\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: \cancel{\dfrac{1}{2}}mv^2 = \cancel{\dfrac{1}{2}}kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: mv^2 = kx^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v^2 = \dfrac{kx^2}{m}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: \sqrt{{v}^{2} } = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13}^{2}}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13} \times 0.13}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times 0.0169}{12}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{13.52}{12}}}}[/tex]
[tex]{\star{\underline{\boxed{\rm{\red{ v \approx 1.1266 \: m/s}}}}}}[/tex]
Hence, the speed of block is 1.1266 m/s.
[tex] \rule{300}{1.5}[/tex]
