A block of mass 0.18 kg is attached to a spring of force-constant 2N/m. The coefficient of friction between the block and the flow is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is V = N/10. Then N is :

Engineering

Physics

Physical Quantities

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Using W – E theorem

$\frac{1}{2} \times m {(u)}^{2} = \frac{1}{2} K {(x)}^{2} + μmg (x)$

$\frac{1}{2} \times (0 .18) u^{2} = \frac{1}{2}$ × 2 × 36 × 10^–4 + 0.1 × 0.18 × 10 × 0.06

$\Rightarrow$ u = 0.4 m/sec.

$\Rightarrow \frac{4}{10} m / \sec .$

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