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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A block whose mass $m$ is $15\; kg$ slides on a horizontal frictionless surface with a constant speed $v$ of $3.7m/s$.It is brought momentarily to rest as it compresses a spring in its path.By what distance $d$ is the spring compressed?The spring constant $k$ is $1500\;N/m$

$\begin{array}{1 1}37cm\\38cm\\36cm\\35cm\end{array} $

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1 Answer

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Answer : 37 cm
The work done by the spring force on the block as the spring is compressed a distance $d$ from its rest state is given by
$W_s=-\large\frac{1}{2}$$kd^2$
The change in the kinetic energy of the block as it is stopped is
$\Delta K =K_f-K_i$
$\;\;\quad= 0-\large\frac{1}{2}$$mv^2$
The work-energy theorem requires that these two quantities be equal.Setting them so
$W_s=\Delta K$
$\Rightarrow -\large\frac{1}{2}$$kd^2$
$\Rightarrow -mv^2$
$d=v\sqrt{\large\frac{m}{k}}$
$\;\;\;=3.7m/s(15kg/1500N/m)^{\large\frac{1}{2}}$
$d=0.37m=37cm$
answered Sep 1, 2014 by sreemathi.v
 

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