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If the acceleration of wedge in the shown arrangement is a ms-2 towards left, then at this instant, acceleration of the block (magnitude only) would be
detailed solution
Correct option is B
If the wedge moves leftward by x, then the block moves down the wedge by 4x, i.e., w.r.t. wedge the block comes down by 4x. So, acceleration of block w.r.t wedge = 4a along the incline plane of wedge. Acceleration of wedge with respect to ground= a, along left. So acceleration of block w.r.t. ground is the vector sum of the two vectors shown in the figure. That isa→BG=a2+(4a)2+2×a×4a×cos(π−α)=(17−8cosα)ams−2Talk to our academic expert!
Similar Questions
Figure shows two blocks, each of mass m. The system is released from rest. If accelerations of blocks
A and B at any instant (not initially) are a1 and a2, respectively, then
.
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