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CBSE Class 10 Answered

A block is released from rest at the top of  an frictionless inclined plane 16 m long. It reaches the bottom  4 seconds later. A second block is projected up the plane from the bottom at an instant the block is released in a such a way that it returns to the bottom simultaneously with first block.  Find the accleration of each block on the inclined plane.  [EXPLAIN WITH PROPER STEPS]
Asked by acv27joy | 22 May, 2018, 07:28: PM
answered-by-expert Expert Answer
we need to use the formula " S = u×t + (1/2)a×t2 " with u =0 because the first  block starts from rest
when it is released from the top of inclined plane.
 
16 = (1/2)×a×4×4 ............(1)
 
above equation gives acceleration a = 2 m/s2
 
Both the blocks will experience same acceleration. Acceleration is positive for the block that slides down.
Acceleration is negative for the block that slides up
Answered by Thiyagarajan K | 19 Jun, 2018, 08:36: PM
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