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PhysicsNLMFrictionMedium2 minPYQ_2023
PhysicsMediumsingle choice

A block of mass5 kgis placed at rest on a table of rough surface. Now, if a force of30 Nis applied in the direction parallel to surface of the table, the block slides through a distance of50 min an interval of time10 s. Coefficient of kinetic friction is (given,g = 10 m s2):

Question diagram: A block of mass 5 kg is placed at rest on a table of rough s

Options:

Answer:
C
Solution:

Forces acting on block is shown below

From the second equation of motion,

s=12at250=12a×102a=1 m s-2

Applying Newton's second law in vertical direction,

R-mg=0R=mg

(where R is the magnitude of normal reaction on the block due to the surface)

If μ is the coefficient of kinetic friction between the block and the surface, applying Newton's second law in horizontal direction,

30-μR=ma30-μmg=ma30-50μ=5μ=12=0.5

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction
2mℹ️ Source: PYQ_2023

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