A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the block is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take g = 10 m/s².

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Solution

Consider the free body diagram.

(a) For the mass of 2 kg, we have:
R₁ − 2g = 0
⇒ R₁ = 2 × 10 = 20
2a + 0.2 R₁ − 12 = 0
⇒ 2a + 0.2 (20) = 12
⇒ 2a = 12 − 4
⇒ a = 4 m/s²

Now,
4a − μR₁ = 0
⇒ 4a = μR₁ = 0.2 (20) = 4
⇒ a₁ = 1 m/s²

The 2 kg block has acceleration 4 m/s² and the 4 kg block has acceleration 1 m/s².

(ii) We have:
R₁ = 2g = 20
Ma = μR₁ = 0
a = 0
And,
Ma + μmg − F = 0
4a + 0.2 × 2 × 10 − 12 = 0
⇒ 4a + 4 = 12
⇒ 4a = 8
⇒ a = 2 m/s²

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