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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2006 · Physics · Laws of Motion

A circular disc with a groove along its diameter is placed...

A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg1 \mathrm{~kg} is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is μ=2/5\mu=2 / 5. The disc has an acceleration of 25 m/s225 \mathrm{~m} / \mathrm{s}^2. Find the acceleration of the block with respect to disc. A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 ~kg is placed as shown. The

Correct answer

10

Step-by-step solution

Normal reaction in vertical direction N1=mgN_1=m g Normal reaction from side to the groove N2=masin⁡37∘N_2=m a \sin 37^{{}^{\circ}} Therefore, acceleration of block with respect to discs a_r=macos⁡37∘−μN1−μN2m\frac{m a \cos 37^{{}^{\circ}}-\mu N_1-\mu N_2}{m}Substituting the values we get, ar=10 m/s2\quad a_r=10 \mathrm{~m} / \mathrm{s}^2
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