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A \( 10 \mathrm{~kg} \) metal block is attached to a spring of spring constant \( 1000 \mathrm{~N} \mathrm{~m}^{-1} \), A block is
displaced from equilibrium position by \( 10 \mathrm{~cm} \) and released. The maximum acceleration of the
block is
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displaced from equilibrium position by \( 10 \mathrm{~cm} \) and released. The maximum acceleration of the
block is
Solution:
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Verified Answer
The correct answer is:
\( 10 \mathrm{~ms}^{-2} \)
Given, mass, $\mathrm{m}=10 \mathrm{~kg}$; spring constant $k=1000 \mathrm{Nm}^{-1}$, displacement $x=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}$
We know $\mathrm{F}=\mathrm{kx}$
Also, $\mathrm{F}=\mathrm{ma}$
$\Rightarrow \mathrm{kx}=\mathrm{ma}$
$\Rightarrow a=\frac{k x}{\mathrm{~m}}=\frac{1000 \times 10 \times 10^{-2}}{10}=10 \mathrm{~ms}^{-2}$
Therefore, maximum acceleration of the block is $10 \mathrm{~ms}^{-2}$
We know $\mathrm{F}=\mathrm{kx}$
Also, $\mathrm{F}=\mathrm{ma}$
$\Rightarrow \mathrm{kx}=\mathrm{ma}$
$\Rightarrow a=\frac{k x}{\mathrm{~m}}=\frac{1000 \times 10 \times 10^{-2}}{10}=10 \mathrm{~ms}^{-2}$
Therefore, maximum acceleration of the block is $10 \mathrm{~ms}^{-2}$
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