A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an angle of 30° with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is $$\mu$$ = 0.2. The difference between the accelerations of the block, in case (B) and case (A) will be: (g = 10 m s$$^{-2}$$)

Case (A) - Push:

$$N_A = mg + F\sin\theta = 5(10) + 20\sin30^\circ = 50 + 10 = 60\text{ N}$$

$$F_{\text{net}, A} = F\cos\theta - \mu N_A = m a_A$$

Case (B) - Pull:

$$N_B = mg - F\sin\theta = 5(10) - 20\sin30^\circ = 50 - 10 = 40\text{ N}$$

$$F_{\text{net}, B} = F\cos\theta - \mu N_B = m a_B$$

$$m(a_B - a_A) = \mu(N_A - N_B)$$

$$5(a_B - a_A) = 0.2(60 - 40) = 0.2 \times 20 = 4$$

$$a_B - a_A = \frac{4}{5} = 0.8\text{ m/s}^{-2}$$