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(D) Case $(A)$ (Pushing): Vertical forces: $N_1 = mg + F \sin 30^o = 5 	imes 10 + 20 	imes 0.5 = 50 + 10 = 60\, N$. Horizontal force: $F_x = F \cos

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$0.8$

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(D) Case $(A)$ (Pushing): Vertical forces: $N_1 = mg + F \sin 30^o = 5 	imes 10 + 20 	imes 0.5 = 50 + 10 = 60\, N$. Horizontal force: $F_x = F \cos 30^o = 20 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3}\, N$. Friction: $f_1 = \mu N_1 = 0.2 	imes 60 = 12\, N$. Acceleration: $a_1 = \frac{F_x - f_1}{m} = \frac{10\sqrt{3} - 12}{5} = 2\sqrt{3} - 2.4 \approx 3.464 - 2.4 = 1.064\, ms^{-2}$. Case $(B)$ (Pulling): Vertical forces: $N_2 = mg - F \sin 30^o = 5 	imes 10 - 20 	imes 0.5 = 50 - 10 = 40\, N$. Horizontal force: $F_x = F \cos 30^o = 10\sqrt{3}\, N$. Friction: $f_2 = \mu N_2 = 0.2 	imes 40 = 8\, N$. Acceleration: $a_2 = \frac{F_x - f_2}{m} = \frac{10\sqrt{3} - 8}{5} = 2\sqrt{3} - 1.6 \approx 3.464 - 1.6 = 1.864\, ms^{-2}$. Difference: $a_2 - a_1 = (2\sqrt{3} - 1.6) - (2\sqrt{3} - 2.4) = 0.8\, ms^{-2}$.

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