CBSE Class 11-science Answered
please solve this
Asked by suchananag2002 | 04 Oct, 2019, 01:29: PM
Expert Answer
Free body diagrams of block of mass m and wedge of mass M are shown in figure.
As seen from figure, Normal contact force between block and wedge is pushing the wedge in horizontal direction
Hence , mg cosα sinα = M×a .....................(1)
where a is acceleration and α is angle of wedge
From eqn.(1), we get acceleration a = (1/2) (m/M) (sin2α) g
Answered by Thiyagarajan K | 04 Oct, 2019, 02:48: PM
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