A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is . Take m=8 kg and M=16 kg. Assume all the surfaces shown in the figure to be frictionless.

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(4) 2/3g. Idea: the wedge recoils, so the block's acceleration down the slope is larger than the fixed-wedge value g sin θ. The standard result is aᵣₑₗ=((M+m)g sin θ)/(M+m sin²θ). Here θ=30°, so sin θ=1/2 and sin²θ=1/4. Numerator: (16+8)g×1/2=12g. Denominator: 16+8×1/4=16+2=18. aᵣₑₗ=(12g)/(18)=2/3g. Sanity check: a fixed wedge would give g sin 30°=0.5g, and 2/3g is duly larger — the wedge sliding out from under the block lets it drop faster.

A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is ______. Take m=8 kg and M=16 kg. Assume all the surfaces shown in the figure to be frictionless.

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