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A $2\ kg$ block slides on a horizontal floor with a speed of $4\ m/s$. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is $15\ N$ and spring constant is $10,000\ N/m$. The spring compresses by ............. $\mathrm{cm}$
$5.5 $
$2.5$
$11$
$8.5$
Solution
Let the block compress the spring by $x$ before stopping.
kinetic energy of the block $=$ ($P.E$ of commpressed spring)$+$work done against friction.
$\frac{1}{2} \times 2 \times {\left( 4 \right)^2} = \frac{1}{2} \times 10,000{x^2} + 15 \times x$
$10,000$${x^2}$$+30x – 32 = 0$
$ \Rightarrow \,5000{x^2} + 15x – 16 = 0$
$\therefore \,x = \frac{{ – 15 \pm \sqrt {{{\left( {15} \right)}^2} – 4 \times \left( {5000} \right)\left( { – 16} \right)} }}{{2 \times 5000}}$
$=0.055m = 5.5 cm.$
