Question Number 165791 by aurpeyz last updated on 08/Feb/22
Answered by ghchoi0707 last updated on 08/Feb/22
![m=Mass of roller coaster (any unit) g=Acceleration by Gravity≒9.8m/s^2 z=Position of roller coaster (m) v_i /v_f =Initial/Final velocity of roller coaster (m/s) E=Mechanical Energy K=Kinetic Energy U=Potential Energy (By Gravity) We have to find v_f . Let [Initial U]=0J. Then ΔU= mgΔz=mg×(−20)=−20mg (∵ U is path independent. In other words, you just have to check only initial and final condition.) as ΔE=0, ΔK=(m/2)×(v_f ^( 2) −v_i ^( 2) )=−ΔU=20mg. as v_i =0, v_f =(√(40g))≒19.8 (m/s).](https://www.tinkutara.com/question/Q165803.png)
$${m}=\mathrm{Mass}\:\mathrm{of}\:\mathrm{roller}\:\mathrm{coaster}\:\left(\mathrm{any}\:\mathrm{unit}\right) \\ $$$${g}=\mathrm{Acceleration}\:\mathrm{by}\:\mathrm{Gravity}\fallingdotseq\mathrm{9}.\mathrm{8m}/\mathrm{s}^{\mathrm{2}} \\ $$$${z}=\mathrm{Position}\:\mathrm{of}\:\mathrm{roller}\:\mathrm{coaster}\:\left(\mathrm{m}\right) \\ $$$${v}_{{i}} /{v}_{{f}} =\mathrm{Initial}/\mathrm{Final}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{roller}\:\mathrm{coaster}\:\left(\mathrm{m}/\mathrm{s}\right) \\ $$$${E}=\mathrm{Mechanical}\:\mathrm{Energy} \\ $$$${K}=\mathrm{Kinetic}\:\mathrm{Energy} \\ $$$${U}=\mathrm{Potential}\:\mathrm{Energy}\:\left(\mathrm{By}\:\mathrm{Gravity}\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find}\:{v}_{{f}} \:. \\ $$$$\mathrm{Let}\:\left[\mathrm{Initial}\:{U}\right]=\mathrm{0J}. \\ $$$$\mathrm{Then}\:\Delta{U}=\:{mg}\Delta{z}={mg}×\left(−\mathrm{20}\right)=−\mathrm{20}{mg} \\ $$$$\left(\because\:{U}\:\mathrm{is}\:\mathrm{path}\:\mathrm{independent}.\right. \\ $$$$\left.\mathrm{In}\:\mathrm{other}\:\mathrm{words},\:\mathrm{you}\:\mathrm{just}\:\mathrm{have}\:\mathrm{to}\:\mathrm{check}\:\mathrm{only}\:\mathrm{initial}\:\mathrm{and}\:\mathrm{final}\:\mathrm{condition}.\right) \\ $$$$\mathrm{as}\:\Delta{E}=\mathrm{0},\:\Delta{K}=\frac{{m}}{\mathrm{2}}×\left({v}_{{f}} ^{\:\:\mathrm{2}} −{v}_{{i}} ^{\:\:\mathrm{2}} \right)=−\Delta{U}=\mathrm{20}{mg}. \\ $$$$\mathrm{as}\:{v}_{{i}} =\mathrm{0},\:{v}_{{f}} =\sqrt{\mathrm{40}{g}}\fallingdotseq\mathrm{19}.\mathrm{8}\:\left(\mathrm{m}/\mathrm{s}\right). \\ $$
Answered by MJS_new last updated on 09/Feb/22
$$\mathrm{ignoring}\:\mathrm{friction}\:\mathrm{and}\:\mathrm{air}\:\mathrm{resistance}\:\mathrm{it}'\mathrm{s}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{free}\:\mathrm{fall}\:\mathrm{of}\:\mathrm{20m}: \\ $$$${s}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:{t}=\sqrt{\frac{\mathrm{2}{s}}{{g}}} \\ $$$${v}={gt}=\sqrt{\mathrm{2}{gs}}=\sqrt{\mathrm{2}×\mathrm{9}.\mathrm{81}×\mathrm{20}}\approx\mathrm{19}.\mathrm{8}\frac{\mathrm{m}}{\mathrm{s}} \\ $$