A constant force produces maximum velocity v on the block connected to the spring of force constant k as shown in the figure. When the force constant of spring becomes $$4k$$,the maximum velocity of the block is (block$$ is at rest when spring is $$relaxed):

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Infinity Learn · Accepted Answer

By work energy theorem;$$Fx1$$−$$12kx12=12mv2$$     .......(1)and$$ $$Fx2$$−$$12k$$′$$x22=12mv$$′$$2$$   .........$$(2)where$$ $$x1$$,$$x2$$ are initial and final extensions and v,t' are initialand final velocities.In both cases, force applied is same, and velocity becomes maximum when $$F=$$$$kx$$. $$(after$$ which the mass will $$decelerate)∴ $$F=kx1=(4k)x2$$⇒ $$x2=x14Substituting$$ in (2):$$Fx14$$−$$12(4K)x142=12mv$$′$$2$$⇒ $$14Fx1$$−$$12kx12=12mv$$′$$2$$     .......(3)Dividing$$ $$(3)/(1), we get:$$14=v$$′$$2v2$$⇒v′$$=v2Hence$$, (c).

A constant force produces maximum velocity v on the block connected to the spring of force constant k as shown in the figure. When the force constant of spring becomes 4k,the maximum velocity of the block is (block is at rest when spring is relaxed):

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