Physics - NLM Question with Solution | TestHub

PhysicsNLMFrictionEasy2 minQB
PhysicsEasynumerical

The figure below shows a block of mass 15 kg, kept on a rough inclined plane of angle 30° and coefficient of static friction equal to 0.5. It is being acted upon by two forces. What should be the minimum value of P (in N) so that the block doesn't slip downwards? [Take g=10 ms-2 and 3=1.7]

Answer:
16.25
Solution:

F+mgsinθ=P+μ mg cosθ

∴5+ 15×10× 1 2 =P= 0.5×15×10× 3 2

∴80=P+ 75 3 2

∴P=80− 75 3 2

∴P= 160−75 3 2

∴P=16.25 N

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction
⏱ 2mℹ️ Source: QB

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