A wedge Y of mass 10 kg has all frictionless surfaces and its inclined surface makes 37° with the horizontal. A block X of mass 2 kg is placed at the highest point of the wedge as shown in the figure. At t=0 the wedge Y is pulled towards the right with a constant force F of 24 N. Taking the block X…

Question

A wedge Y of mass 10 kg has all frictionless surfaces and its inclined surface makes 37° with the horizontal. A block X of mass 2 kg is placed at the highest point of the wedge as shown in the figure. At t=0 the wedge Y is pulled towards the right with a constant force F of 24 N. Taking the block X to be at rest at t=0, the time taken by it to slide down 8.8 m along the slope, while Y is on the move, is s. (Take tan 37°=3/4 and g=10 m/s²)

Accepted Answer

(1) 2. Idea: the wedge accelerates, so the block's motion has to be written relative to it and the two equations solved together. Let A be the wedge's acceleration and aᵣ the block's acceleration down the slope relative to the wedge, with sin 37°=0.6, cos 37°=0.8. Block, perpendicular and along: eliminating N gives 7.5-A=1.25 aᵣ. Wedge: 10A=24-0.6N with N=25-1.5 aᵣ, so A=0.9+0.09 aᵣ. Substituting: 1.25 aᵣ=6.6-0.09 aᵣ, so aᵣ=4.93 m/s². 8.8=1/2(4.93)t² gives t²=3.57 and t=1.89≈2 s. Check the sense: holding the block still on a frictionless 37° slope would need g tan 37°=7.5 m/s², and the wedge gets nowhere near that, so the block does slide down.

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