What is the acceleration of the block and the trolley system shown in figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? Take g = 10 ms^–2. Neglect the mass of the string. 

As the block and the trolley are connected together by a string of fixed length, both will have same acceleration a. 

Applying Newton’s second law to the motion of the block, 

                                30 – T = 3a                 … (i) 

Applying second law to the motion of the trolley, 

                                    T – f_k = 20a

But f_k = μ_k  R = μ_k  mg = 0.04 × 20 × 10 = 8 N 

∴    T – 8 = 20a                                … (ii) 

Adding (i) and (ii), we get 

22 = 23a or a = 22/23 = 0.96 ms^–2

From (i), 30 – T = 3 × 0.96

T = 30 – 2.88 = 27.12 N