A block of mass 0.18 kg is attached to a spr | vedclass

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Question

Given

Mass of block, $m =0.18 kg$

Spring Constant, $k =2 N / m$

coefficient of friction, $\mu=0.1$

Distance travelled $x =0.06 m$

Initi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Given

Mass of block, $m =0.18 kg$

Spring Constant, $k =2 N / m$

coefficient of friction, $\mu=0.1$

Distance travelled $x =0.06 m$

Initial velocity, $v =\frac{ N }{10} m / s$

To Find

N

Procedure

The forces acting on the spring, during its motion are, weight mg , normal force N (=mg) and friction, $f (=\mu N =\mu mg )$.

$	herefore$ Work done by friction $= fx =\mu mgx .$. (1)

Potential Energy in the spring $=\frac{1}{2} kx ^2 ightarrow(2)$

Initial kinetic energy $=\frac{1}{2} mv ^2 ightarrow(3)$

The initial kinetic energy is stored as potential energy in the spring, and as energy lost against friction.

$	herefore \frac{1}{2} mv ^2=\frac{1}{2} kx ^2+\mu mgx$

$\Rightarrow V =\sqrt{\frac{ kx ^2}{m}+2 \mu gx }$

$=\sqrt{\frac{2 	imes 0.06^2}{0.18}+2 	imes 0.1 	imes 10 	imes 0.06}$

$\Rightarrow v =0.4$

Since, $v=\frac{N}{10}=0.4$, we see $N=4$

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