Question 5.3: Additional Answer Options

2022 - May/June - Paper 1 - Question 5

Or

Option 2

\left.\begin{array}{rl}\mathrm{W}_{\mathrm{nc}} & =\Delta\mathrm{K}+\Delta U\\ \mathrm{W}_{\mathrm{f}} & =\Delta\mathrm{K}+mg\left(\mathrm{h}_{\mathrm{f}}-\mathrm{h}_{\mathrm{i}}\right)\\ \mathrm{f}\Delta\mathrm{x}\cos\theta & =1/2mv_{\mathrm{f}}^2-1/2mv_{\mathrm{i}}^2+mgh_{\mathrm{f}}-mgh_{\mathrm{i}}\end{array}\right\}Anyone

31000\Delta x\cos180^{\circ}=1/2(30000)\left(0^2-33^2\right)+30000(9,8)\left(\Delta x\sin28^{\circ}-0\right)

\Delta x=96,64m

Or

Option 3

\left.\begin{array}{rl}W_{\text{net }} & =\Delta K\\ W_{w}+W_{f} & =1/2mv_{f}^2-1/2mv_{i}^2\\ -\Delta E_{p}+W_{f} & =1/2mv_{f}^2-1/2mv_{i}^2\\ -mg\left(h_{f}-h_{i}\right)+W_{f} & =1/2mv_{f}^2-1/2mv_{i}^2\end{array}\right\}Anyone

-(30000)(9,8)\left(\Delta x\sin28^{\circ}-0\right)+(31000)\Delta x\cos180^{\circ}=1/2(30000)\left(0^2-33^2\right)

\Delta x=96,64m

Or

Option 4

\left.\begin{array}{c}W_{net}=\Delta K\\ W_{w}+W_{f}=1/2mv_{f}^2-1/2mv_{i}^2\\ mg\sin\theta\Delta x\cos\theta+W_{f}=1/2mv_{f}^2-1/2mv_{i}^2\end{array}\right\}Anyone

(30000)(9,8)\sin28^{\circ}\Delta x\cos180^{\circ}+(31000)\Delta x\cos180^{\circ}

=1/2(30000)\left(0^2-33^2\right)

\Delta x=96,64m

Or

Option 5

\mathrm{F}_{\mathrm{net}}=\mathrm{ma}

F_{net}=F_{w//}+f

=(30000)(9,8)\sin28^{\circ}+31000

=169024,64\mathrm{~N}

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

169024,64\Delta x\cos180^{\circ}=1/2(30000)\left(0^2-33^2\right)

\Delta x=9664\mathrm{~m}

Related subjects & topics
Grade 12 Physical Sciences
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