A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley as shown below.
The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g = 10 ms⁻².)

Solution

Physics sir · 3 steps

Treat both blocks as one system: net force is the hanging weight minus friction, shared by 44 kg.

1. Forces along motion. The 4 kg weight pulls the system with 4 × 10 = 40 N. Kinetic friction on the 40 kg block opposes this with μ m g = 0.02 × 40 × 10 = 8 N.
2. Whole system. The string is inextensible, so both blocks share the same acceleration. The tension is internal to the system and cancels, so the total mass that accelerates is 40 + 4 = 44 kg.

   a = (m₂ g - μ m₁ g)/(m₁ + m₂) = (40 - 8)/44 = 32/44 m s⁻²

3. So. a = 8/11 m s⁻², which is option (d).

The idea to keep. For connected blocks, net external force divided by the total moving mass gives the common acceleration. Tension never enters, but every moving mass does.

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