
Ncos60∘=Ma1=16a1
⇒N=32a1
⊥ to incline
N=8gcos30∘−8a1sin30∘⇒32a1=443g−4a1
⇒a2=93g
Along incline
8gsin30∘+8a1cos30∘=ma2=8a2
a2=g×21+93g⋅23=32g
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:
Given m=8kg,M=16kg
Assume all the surfaces shown in the figure to be frictionless.

Held on 1 Sept 2021 · Verified 6 Jul 2026.
53g
34g
56g
32g
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