One end of a spring of force constant 150 tex | vedclass

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(C) First,convert the force constant to $SI$ units:$k = 150 	ext{ dyne cm}^{-1} = 150 	imes 10^{-5} 	ext{ N} / 10^{-2} 	ext{ m} = 0.15 	ext{ N m}

Vedclass · Accepted Answer

$6$

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(C) First,convert the force constant to $SI$ units:$k = 150 	ext{ dyne cm}^{-1} = 150 	imes 10^{-5} 	ext{ N} / 10^{-2} 	ext{ m} = 0.15 	ext{ N m}^{-1}$.Let the block move a distance $x$ before coming to rest. For the block to travel only in one direction,it must stop at a point where the spring force is less than or equal to the maximum static friction.Using the work-energy theorem: $W_{	ext{spring}} + W_{	ext{friction}} = \Delta K$$-\frac{1}{2} k x^2 - \mu m g x = 0 - \frac{1}{2} m v^2$$v^2 = \frac{k x^2}{m} + 2 \mu g x$For the block to not return,the spring force at the maximum extension $x$ must be less than or equal to the limiting friction: $k x \leq \mu m g$.$x \leq \frac{\mu m g}{k} = \frac{0.3 	imes 0.2 	imes 10}{0.15} = \frac{0.6}{0.15} = 4 	ext{ m}$.Substituting $x = 4 	ext{ m}$ into the energy equation:$v^2 = \frac{0.15 	imes 4^2}{0.2} + 2 	imes 0.3 	imes 10 	imes 4$$v^2 = \frac{0.15 	imes 16}{0.2} + 24 = 12 + 24 = 36$$v = 6 	ext{ ms}^{-1}$.

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