In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are…

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°=34)

  1. 9 kg
  2. 4.5 kg
  3. 6.5 kg
  4. 2.25 kg

Solution

For M block:

10gsin53°-μ(10g) cos53°-T=10×2

T=80-15-20

T=45 N

For m block:

Tmg sin 37°-μ mg cos 37°=m×2

45=10 m

m=4.5 kg

Asked in: JEE Main 2024 (31 Jan Shift 1)

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