Question 3.2.1: Additional Answer Options

2021 - November - Paper 1 - Question 3

Or

Option 2

\left.\begin{array}{c}\left(\mathrm{E}_{\text{mech}}\right)_{200\mathrm{m}}=\left(\mathrm{E}_{\text{mech }}\right)_{\text{bottom}}\\ \left(\mathrm{E}_{\mathrm{P}}+\mathrm{E}_{\mathrm{K}}\right)_{200\mathrm{m}}=\left(\mathrm{E}_{\mathrm{P}}+\mathrm{E}_{\mathrm{K}}\right)_{\text{bottom}}\\ \left(\mathrm{mgh}+1/2\mathrm{mv}^2\right)_{200\mathrm{m}}=\left(\mathrm{mgh}+1/2\mathrm{mv}^2\right)_{\text{bottom }}\end{array}\right\}Anyone

m(9,8)(200)+1/2m\left(v^2\right)=0+1/2m(62,68)^2

\mathrm{v}_{\mathrm{i}}=2,96\mathrm{~m}\cdot\mathrm{s}^{-1}

Or

Option 3

\left.\begin{array}{l}W_{nc}=\Delta E_{p}+\Delta E_{k}\\ 0=mg\left(h_{f}-h_{i}\right)+1/2m\left(v_{f}^2-v_{i}^2\right)\end{array}\right\}Anyone

0=m(9,8)(0-200)+1/2m\left(62,68^2-v_{i}^2\right)

\mathrm{v}_{\mathrm{i}}=2,96\mathrm{~m}\cdot\mathrm{s}^{-1}

Or

Option 4

\left.\begin{array}{l}W_{\text{net }}=\Delta E_{k}\\ F_{net}\Delta x\cos\theta=1/2m\left(v_{f}^2-v_{i}^2\right)\\ mg\Delta x\cos\theta=1/2m\left(v_{f}^2-v_{i}^2\right)\end{array}\right\}Anyone

m(9,8)(200)=+1/2m\left(62,68^2-v_{i}^2\right)

v_{i}=2,96\mathrm{~m}\cdot\mathrm{s}^{-1}

Related subjects & topics
Grade 12 Physical Sciences
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