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第 1 步Let us first declare some variables:h: the ramp's elevation (20 \mathrm{~m})\theta : the ramp's inclination (30 degrees)\mathrm{m} : the weight of the block (5 \mathrm{~kg})g: gravity's acceleration \left(9.81 \mathrm{~m} / \mathrm{s}^{\wedge} 2\right).第 2 步\mathrm{U}=\mathrm{mgh}The block at the bottom of the ramp has the kinetic energy:\mathrm{K}=(1 / 2) \mathrm{mv} \mathrm{v}^{\wedge}where v is the velocity of the block at the bottom of the ramp.第 3 步m g h=(1 / 2) m v^{\wedge} 2 v=\operatorname{sqr}(2 g h)Substituting the values yields:v=\operatorname{sqr}\left(2{ }^{*} 9.81 \mathrm{~m} / \mathrm{s}^{\wedge} 2 \times 20 \mathrm{~m}^{*} \sin (30\right. degrees \left.)\right)v=14.01 \mathrm{~m} / \mathrm{s}As a result, the block's velocity at the bottom of the ramp is 14.01 \mathrm{~m} / \mathrm{s}.Final answerAs a result, the block's velocity at the bottom of the ramp is 14.01 \mathrm{~m} / \mathrm{s}. ...