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 / s², is: (take g=10 m / s² and tan 37°=3 / 4)

Solution

Physics sir · 4 steps

Write one force equation for each block along its own slope, then use the string tension to link them.

1. Directions. M slides down the 53° face, so m is pulled up the 37° face with the same a = 2 m s⁻². Friction opposes the motion, so it acts up the slope on M and down the slope on m. With tan 37° = 3/4, we have sin 37° = cos 53° = 0.6 and cos 37° = sin 53° = 0.8.
2. Block M. The gravity component along the slope is Mg sin 53° = 100 × 0.8 = 80 N. Friction is μ Mg cos 53° = 0.25 × 100 × 0.6 = 15 N. So 80 - 15 - T = 10 × 2, which gives T = 45 N.

   T - mg sin 37° - μ mg cos 37° = ma

3. Block m. Put in the numbers: 45 - 6m - 0.25 × 8m = 2m, so 45 - 6m - 2m = 2m. That gives 45 = 10m.
4. So. m = 4.5 kg, which is option (a).

The idea to keep. On a double incline, treat each block on its own slope. Friction always opposes that block's motion, and the tension is the only force the two blocks share.

Answer from the NTA final answer key. Solution by PrepWiser.