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Question Number 165791 by aurpeyz last updated on 08/Feb/22

Answered by ghchoi0707 last updated on 08/Feb/22

$$m=\mathrm{Mass}\mathrm{of}\mathrm{roller}\mathrm{coaster}\left(\mathrm{any}\mathrm{unit}\right)$$$$g=\mathrm{Acceleration}\mathrm{by}\mathrm{Gravity}\fallingdotseq 9.8\mathrm{m}/{\mathrm{s}}^2$$$$z=\mathrm{Position}\mathrm{of}\mathrm{roller}\mathrm{coaster}\left(\mathrm{m}\right)$$$$v_i/v_f=\mathrm{Initial}/\mathrm{Final}\mathrm{velocity}\mathrm{of}\mathrm{roller}\mathrm{coaster}\left(\mathrm{m}/\mathrm{s}\right)$$$$E=\mathrm{Mechanical}\mathrm{Energy}$$$$K=\mathrm{Kinetic}\mathrm{Energy}$$$$U=\mathrm{Potential}\mathrm{Energy}\left(\mathrm{By}\mathrm{Gravity}\right)$$$$\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}{v}_f.$$$$\mathrm{Let}\left[\mathrm{Initial}U\right]=0\mathrm{J}.$$$$\mathrm{Then}\mathrm{\Delta }U=mg\mathrm{\Delta }z=mg\times (-20)=-20mg$$$$(\because U\mathrm{is}\mathrm{path}\mathrm{independent}.$$$$\mathrm{In}\mathrm{other}\mathrm{words},\mathrm{you}\mathrm{just}\mathrm{have}\mathrm{to}\mathrm{check}\mathrm{only}\mathrm{initial}\mathrm{and}\mathrm{final}\mathrm{condition}.)$$$$\mathrm{as}\mathrm{\Delta }E=0,\mathrm{\Delta }K=\frac{m}{2}\times (v_f^2-v_i^2)=-\mathrm{\Delta }U=20mg.$$$$\mathrm{as}{v}_i=0,v_f=\sqrt{40g}\fallingdotseq 19.8\left(\mathrm{m}/\mathrm{s}\right).$$

Answered by MJS_new last updated on 09/Feb/22

$$\mathrm{ignoring}\mathrm{friction}\mathrm{and}\mathrm{air}\mathrm{resistance}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{equal}$$$$\mathrm{to}\mathrm{a}\mathrm{free}\mathrm{fall}\mathrm{of}20\mathrm{m}:$$$$s=\frac{{gt}^2}{2}\Rightarrow t=\sqrt{\frac{2s}{g}}$$$$v=gt=\sqrt{2gs}=\sqrt{2\times 9.81\times 20}\approx 19.8\frac{\mathrm{m}}{\mathrm{s}}$$

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