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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A $1.0 \;kg$ block collides with a horizontal weightless spring of spring constant $2.0N/m$. The block compresses the spring $4.0\;m$ from the rest position.Assuming that the coefficient of kinetic friction between the block and horizontal surface of $0.25,$ what was the speed of the block at the instant of collision

$ a)\; 7.18 m/s \\ b)\; 6.23 m/s \\ c)\; 5.21 m/s \\ d)\; 4.52 m/s $

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1 Answer

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Let $'v'$ be the velocity of block at the time of collision.
$KE$ of block=(Elastic energy store in spring )+ energy spent in doing work against friction
$\large\frac{1}{2}$$mv^2=\large\frac{1}{2}$$kx^2 +\mu _k mg x$
$\large \frac{1}{2} $$\times 1 \times v^2=\large\frac{1}{2} $$\times 2 \times 4^2+(.25) \times 1 \times 9.8 \times 4$
$\large\frac{1}{2}$$v^2=16+9.8$
$v^2=2 \times 25.8$
$v=7.18 \;m/s$
Hence  a is the correct answer. 

 

answered Aug 6, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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