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A block of mass $m=25 \mathrm{~kg}$ sliding on a smooth horizontal surface with a velocity $v=3 \mathrm{~ms}^{-1}$ meets the spring of spring constant $k=100 \mathrm{~N} / \mathrm{m}$ fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as it returns to the original position respectively are
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The correct answer is:
$1.5 \mathrm{~m},-3 \mathrm{~ms}^{-1}$
When block strikes the spring, the kinetic energy of block converts into potential energy of spring ie,
$\frac{1}{2} m v^2=\frac{1}{2} k x^2$
where $x$ is compression in spring.
$x=\sqrt{\frac{m v^2}{k}}$
$=\sqrt{\frac{25 \times(3)^2}{100}}=\sqrt{\frac{9}{4}}=\frac{3}{2}=1.5 \mathrm{~m}$
When block returns to the original position, again potential energy converts into kinetic energy of the block, so velocity of the block is same as before but its sign changes as it goes to mean position.
Hence, $\quad v=-3 \mathrm{~ms}^{-1}$.
$\frac{1}{2} m v^2=\frac{1}{2} k x^2$
where $x$ is compression in spring.
$x=\sqrt{\frac{m v^2}{k}}$
$=\sqrt{\frac{25 \times(3)^2}{100}}=\sqrt{\frac{9}{4}}=\frac{3}{2}=1.5 \mathrm{~m}$
When block returns to the original position, again potential energy converts into kinetic energy of the block, so velocity of the block is same as before but its sign changes as it goes to mean position.
Hence, $\quad v=-3 \mathrm{~ms}^{-1}$.
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