A block of weight W moving with velocity v on | vedclass

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Question

(C) According to the law of conservation of energy,the kinetic energy lost by the block is equal to the potential energy gained by the spring.Let $d$

Vedclass · Accepted Answer

$v\sqrt{\frac{W}{kg}}$

Vedclass · Answer

(C) According to the law of conservation of energy,the kinetic energy lost by the block is equal to the potential energy gained by the spring.Let $d$ be the maximum compression in the spring.The kinetic energy of the block is $K = \frac{1}{2}mv^2$,where $m = \frac{W}{g}$ is the mass of the block.The potential energy stored in the spring at maximum compression is $U = \frac{1}{2}kd^2$.Equating the two: $\frac{1}{2}mv^2 = \frac{1}{2}kd^2$.Substituting $m = \frac{W}{g}$:$\frac{1}{2} \left( \frac{W}{g} \right) v^2 = \frac{1}{2}kd^2$.Solving for $d$:$d^2 = \frac{Wv^2}{kg}$.Therefore,$d = v\sqrt{\frac{W}{kg}}$.

$A$ block of weight $W$ moving with velocity $v$ on a frictionless horizontal surface hits a spring of force constant $k$. The maximum compression in the spring will be at a distance of .........

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