As shown in the figure, two blocks are in contact sliding down an inclined surface of inclination 30^ . The friction coefficient between the block of mass 2.0 kg and the incline is ₁ = 0.20 , and that between the block of mass 4.0 kg and the incline is ₂ = 0.30 . Determine the acceleration of the 2.0 kg block. Take g = 10 m/s ^2 .

Step-by-step solution

Let us first calculate the acceleration of each block as if they were sliding independently. For the 2.0 kg block (upper block): a₁ = g( 30^ - ₁ 30^ ) a₁ = 10 (0.5 - 0.20 3 2 ) = 5 - 3 3.27 m/s ^2 For the 4.0 kg block (lower block): a₂ = g( 30^ - ₂ 30^ ) a₂ = 10 (0.5 - 0.30 3 2 ) = 5 - 1.5 3 2.40 m/s ^2 Since a₁ > a₂ , the upper 2.0 kg block will tend to slide faster than the lower 4.0 kg block. Because it is behind the 4.0 kg block, it will push against it, and both blocks will move together with a common accelera