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(A) By the law of conservation of mechanical energy,the potential energy lost by the particle is equal to the elastic potential energy gained by the s

Vedclass · Accepted Answer

$3.96$

Vedclass · Answer

(A) By the law of conservation of mechanical energy,the potential energy lost by the particle is equal to the elastic potential energy gained by the spring.$mg(h + x) = \frac{1}{2} kx^2$For the first case: $m = 0.1 \; kg$,$h_1 = 0.24 \; m$,$x_1 = 0.01 \; m$.$mg(0.24 + 0.01) = \frac{1}{2} k(0.01)^2$  --- $(1)$For the second case: $h_2 = h$,$x_2 = 0.04 \; m$.$mg(h + 0.04) = \frac{1}{2} k(0.04)^2$  --- $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{h + 0.04}{0.24 + 0.01} = \frac{(0.04)^2}{(0.01)^2}$$\frac{h + 0.04}{0.25} = \frac{0.0016}{0.0001} = 16$$h + 0.04 = 16 	imes 0.25 = 4$$h = 4 - 0.04 = 3.96 \; m$

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