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 potential energy of the spring. By the law of conservation

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 potential energy of the spring. By the law of conservation of energy:$\frac{1}{2} m v^2 = \frac{1}{2} k x^2$where $x$ is the maximum compression in the spring.$x = \sqrt{\frac{m v^2}{k}} = \sqrt{\frac{25 	imes (3)^2}{100}} = \sqrt{\frac{225}{100}} = \frac{15}{10} = 1.5 \ m$.When the block returns to the original position,the potential energy stored in the spring is completely 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.Thus,the velocity of the block is $v = -3 \ ms^{-1}$.

$X = x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

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