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PhysicsNLMFrictionHard2 minPYQ_2024
PhysicsHardsingle choice

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F→1 is the force required to just move the block up the inclined plane and F→2 is the force required to just prevent the block from sliding down, then the value of F→1−F→2 is: [Use g=10 m s-2]

Options:

Answer:
B
Solution:

Limiting friction force is, fmax=μmg cosθ

=0.1×50×32

=2.53 N

For first case:

We can write, F1=mg sinθ+fmax

=25+2.53

For second case:

F2=mg sinθ−fmax

=25−2.53

∴F1−F2=53 N

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction
⏱ 2mℹ️ Source: PYQ_2024

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