Question 5.3: Additional Answer Options

2022 - November - Paper 1 - Question 5

Or

Option 2

\mathrm{W}_{\mathrm{Fg}//}=\mathrm{F}_{\mathrm{g}//}\Delta\mathrm{x}\cos

=(mg\sin\theta)\Delta x\cos\theta

=(12)(9,8)\left(\frac{4,5}{\Delta x}\right)\Delta x\cos180^{\circ}

=-529,2\mathrm{~J}

\mathrm{W}_{\text{net }}=\Delta\mathrm{Ek}

=1/2mv_{f}^2-1/2mv_{i}^2

=1/2(12)(2.25)^2-0^2

=30,375\mathrm{~J}

W_{\text{net }}=W_{\mathrm{nc}}+W_{c}

30,375=\mathrm{W}_{\mathrm{nc}}+(-529,2)

\mathrm{W}_{\mathrm{nc}}=559,575\mathrm{J}(559,58\mathrm{J})

Or

Option 3

\operatorname{Sin}\theta=\frac{4,5}{\Delta x}

\Delta x=\frac{4,5}{\sin\theta}

\left.\begin{array}{rl}\mathrm{W}_{\text{net }} & =\Delta\mathrm{E}_{\mathrm{k}}\\ \mathrm{W}_{\mathrm{F}}+\mathrm{W}_{\mathrm{f}}+\mathrm{W}_{\mathrm{w}} & =\Delta\mathrm{E}_{\mathrm{k}}\\ \mathrm{W}_{\mathrm{nc}}+(\mathrm{mg}\sin\theta)\Delta\mathrm{x}(\cos\beta) & =1/2mv_{\mathrm{f}}^2-1/2\mathrm{mv}_{\mathrm{i}}^2\end{array}\right\}\begin{gathered}\text{ }\\ \text{ }\end{gathered}Anyone

\mathrm{W}_{\mathrm{nc}}+(12)(9,8)\left(\frac{4,5}{\sin\theta}\right)\sin\theta\cos180^{\circ}={1/2(12)(2,25)^2}-0^2

W_{\mathrm{nc}}=559,58\mathrm{~J}

Related subjects & topics
Grade 12 Physical Sciences
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