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PhysicsNLMFrictionMedium2 minQB

PhysicsMediuminteger

In the figure shown if system starts at t = 0, at what time (in sec) 10 kg block reaches to end of table ? (F is external force)

Answer:

Solution:

Limiting friction on the 10 kg block: $f_{ma x} = 0.5 \times 100 = 50 N$ .

Net driving force = $(50 + 100) - 50 = 100 N$ on total 15 kg: $a = \frac{100}{15} = \frac{20}{3} m s^{- 2}$ .

$\frac{40}{3} = \frac{1}{2} \cdot \frac{20}{3} t^{2} \Rightarrow t = 2 s$ . Answer: 2.

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction

⏱ 2mℹ️ Source: QB

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