A block of mass m=25 kg sliding on a smooth | vedclass

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(A) When the block strikes the spring,the kinetic energy of the block is converted into the elastic potential energy of the spring. According to the l

Vedclass · Accepted Answer

$1.5 \ m, -3 \ ms^{-1}$

Vedclass · Answer

(A) When the block strikes the spring,the kinetic energy of the block is converted into the elastic potential energy of the spring. According to the law of conservation of energy:$\frac{1}{2} mv^2 = \frac{1}{2} kx^2$where $x$ is the maximum compression in the spring.Solving for $x$:$x = \sqrt{\frac{mv^2}{k}} = \sqrt{\frac{25 	imes (3)^2}{100}} = \sqrt{\frac{225}{100}} = \sqrt{2.25} = 1.5 \ m$When the block returns to its original position,the potential energy stored in the spring is fully converted back into the kinetic energy of the block. Since the surface is smooth (no friction),energy is conserved. The magnitude of the velocity remains the same,but the direction is reversed as the block moves away from the spring.Therefore,the velocity of the block is $v = -3 \ ms^{-1}$.

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