Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04 , the acceleration of the system in ms⁻² is :
(Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :

Solution

Physics sir · 3 steps

Treat block and trolley as one body: net pull is weight minus friction, shared by all 26 kg.

1. Forces along the string. The hanging 6 kg block pulls the system with its weight, 60 N. Kinetic friction on the 20 kg trolley acts backwards, fₖ = μ m g = 0.04 × 20 × 10 = 8 N.
2. Whole system. The string is light and inextensible, so both bodies move with the same acceleration a. Tension is internal to the system and cancels out.

   a = (m₂ g - μ m₁ g)/(m₁ + m₂) = (60 - 8)/(20 + 6) = 52/26 m s⁻²

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

The idea to keep. For connected bodies, the net external force along the string divided by the total moving mass gives the common acceleration. Friction only depends on the mass pressing on the rough surface.

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