In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of fricti

Question

In the given arrangement of a doubly inclined plane two blocks of masses $M$ and $m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $0.25$ . The value of $m$ , for which $M = 10 kg$ will move down with an acceleration of $2 m s^{- 2}$ , is: (take $g = 10 m s^{- 2}$ and $\tan 37 ° = \frac{3}{4}$ )

TestHub · Accepted Answer

For M block: 10 g sin 53 ° - μ ( 10 g ) cos 53 ° - T = 10 × 2 ⇒ T = 80 - 15 - 20 ⇒ T = 45 N For m block: T – m g sin 37 ° - μ m g cos 37 ° = m × 2 ⇒ 45 = 10 m ⇒ m = 4 . 5 kg

Physics - NLM Question with Solution | TestHub

Options:

Doubts & Discussion