A block of mass 2kg is moving with speed v0 towards a massless unstretched spring (K=10N/m). It is found that for maximum speed v0=6.4m/s, the block compresses the spring, stops at that position and does not return back. Friction coefficient at surface is μ=1x. Value of x is

For the block to stop, μmg=kxμ=kxmg=10×x2×10=x212mV2=12Kx2+μmg(1+x) ⇒122×6.4=1210x2+x22×10(1+x) ⇒x=0.4=25So, μ=x2=15

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A block of mass 2kg is moving with speed $v_{0}$ towards a massless unstretched spring $(K = 10 N / m) .$ It is found that for maximum speed $v_{0} = \sqrt{6.4} m / s,$ the block compresses the spring, stops at that position and does not return back. Friction coefficient at surface is $μ = \frac{1}{x} .$ Value of x is

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$F o r t h e b l o c k t o s t o p,$ $μ m g = k x$

$μ = \frac{k x}{m g} = \frac{10 \times x}{2 \times 10} = \frac{x}{2}$

$\frac{1}{2} m V^{2} = \frac{1}{2} K x^{2} + μ m g (1 + x)$

$\Rightarrow \frac{1}{2} 2 \times 6.4 = \frac{1}{2} 10 x^{2} + \frac{x}{2} 2 \times 10 (1 + x)$

$\Rightarrow x = 0.4 = \frac{2}{5}$

So, $μ = \frac{x}{2} = \frac{1}{5}$

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