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PhysicsNLMMiscellaneousMedium2 minQB

PhysicsMediuminteger

A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove in contact is $μ = 2/5$ . The disc has an acceleration of $25 m / s^{2}$ . Find the acceleration of the block with respect to disc.

Answer:

Solution:

ma $cos 37 - μ$ ma $sin 37 - μ mg = ma_{0}$

$25 \times \frac{4}{5} - \frac{2}{5} \times 25 \times \frac{3}{5} - \frac{2}{5} \times 10 = a_{0}$ $20 - 6 - 4 = 90$

$a_{0} = 10 m / s^{2}$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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Physics - NLM Question with Solution | TestHub

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