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PhysicsNLMFrictionMedium2 minQB

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For shown situation the acceleration of block is represented by :- (Here t is time in second) ( $g = 10 m s^{- 2}$ )

Answer:

Solution:

Motion starts when $2 t = μ_{s} m g = 0.4 \times 10 \times 10 = 40 N$ , i.e. at t = 20 s; so a = 0 till then.

Just after start: $a = \frac{40 - 30}{10} = 1 m s^{- 2}$ (kinetic friction $= 0.3 \times 10 \times 10 = 30 N$ ).

For t > 20 s: $a = \frac{2 t - 30}{10} = \frac{t}{5} - 3$ , slope $= \frac{1}{5} = tan^{- 1} (0.2)$ . Hence (B).

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction

⏱ 2mℹ️ Source: QB

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