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(D) Let the total distance travelled by the block along the incline be $d = 1 \, m$. According to the law of conservation of mechanical energy,the los

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$5$

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(D) Let the total distance travelled by the block along the incline be $d = 1 \, m$. According to the law of conservation of mechanical energy,the loss in gravitational potential energy of the block is equal to the gain in elastic potential energy of the spring. Let $d$ be the total distance travelled by the block along the incline. The vertical height dropped by the block is $h = d \sin 45^{\circ}$. The block comes to rest after compressing the spring by $x = 1 \, m$. Thus,the total distance travelled by the block is $d = (d_{initial} + x)$. However,the problem asks for the total distance travelled along the incline. Let this distance be $d$. Initial potential energy $U_i = mgd \sin 45^{\circ}$. Final potential energy $U_f = \frac{1}{2} k x^2$. Equating $U_i = U_f$: $mgd \sin 45^{\circ} = \frac{1}{2} k x^2$ $(\sqrt{2}) 	imes 10 	imes d 	imes \frac{1}{\sqrt{2}} = \frac{1}{2} 	imes 100 	imes (1)^2$ $10d = 50$ $d = 5 \, m$.

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