A block of mass m initially at rest is droppe | vedclass

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(B) According to the law of conservation of energy,the total mechanical energy of the system remains constant.Let the initial position of the block be

Vedclass · Accepted Answer

$mg(h + x) = \frac{1}{2}kx^2$

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(B) According to the law of conservation of energy,the total mechanical energy of the system remains constant.Let the initial position of the block be the reference level for gravitational potential energy $(U_g = 0)$.When the block is at height $h$ above the spring,its total energy is $E_i = 0$ (since it is at rest and at the reference level).When the spring is compressed by a maximum distance $x$,the block comes to rest momentarily at the lowest point.The total vertical displacement of the block from its initial position is $(h + x)$.The change in gravitational potential energy is $\Delta U_g = -mg(h + x)$.The elastic potential energy stored in the spring is $U_s = \frac{1}{2}kx^2$.Since the total energy is conserved,the loss in gravitational potential energy is equal to the gain in elastic potential energy:$mg(h + x) = \frac{1}{2}kx^2$.

$A$ block of mass $m$ initially at rest is dropped from a height $h$ onto a spring of force constant $k$. If the maximum compression in the spring is $x$,then:

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