Additional Answer Options: 5.5

5.1

The total mechanical energy/sum of kinetic and gravitational potential energy in a closed/isolated system is constant (conserved).   

2 marks

(2 x 1 mark)

5.2

\left.\begin{array}{l}E_{\text{MECHP }}=E_{\text{MECH }} \\ \left(E_P+E_K\right)_P=\left(E_P+E_K\right)_Q \\ W_{\text{net }}=\Delta E_K \\ W_{\text{con }}=\Delta E_K \\ \Delta E_K+\Delta E_P=0\end{array}\right\}Anyone  

\left(m g h+\frac{1}{2} m v^2\right)_P=\left(m g h+\frac{1}{2} m v^2\right)_Q 

50(9,8)(3)+0 v=0+1 / 2(50) v^2   

v=7,67\mathrm{m}\cdot\mathrm{s}^{-1}\left(7,668\mathrm{m}\cdot\mathrm{s}^{-1}\right)  

4 marks

(4 x 1 mark)

5.3

OR

3 marks

(3 x 1 mark)

5.4

\left.\begin{array}{rl}f_{\mathrm{k}} & =\mu_{\mathrm{k}}\mathrm{N} \\ & =\mu_{\mathrm{k}}(\mathrm{mg}\cos\theta)\end{array}\right\}Anyone  

=0,08(50 \times 9,8) \cos 30^{\circ}  

=33,95(33,948) \mathrm{N}  

3 marks

(3 x 1 mark)

5.5

Option 1:

\left.\begin{array}{l}\begin{aligned} & W=F_{\text{net }}\Delta x\cos\theta \\ & W_{\text{net }}=W_f+W_W+W_N \\ & W_{\text{net }}=W_f+\left(-\Delta E_P\right)+W_N \\ & W_{\text{net }}=f_k\Delta x\cos180^{\circ}+mg\sin\theta\Delta x\cos0+0 \\ & W_{\text{net }}=\Delta E_K/\Delta K\end{aligned}\end{array}\right\}Anyone  

\left.W_{\text{net }}=[33,948)(5)(-1)\right]+\left[(50)(9,8)(5)\sin30^{\circ}+0\right]   

=1055,26(1055,259) 

1055,259=1 / 2(50)\left(v_f^2-7,668^2\right)  

\mathrm{v}_{\mathrm{f}}=10,05 \mathrm{~m} \cdot \mathrm{s}^{-1}  

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5 marks

(5 x 1 mark)