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 $$g = 10$$ m s$$^{-2}$$.)

Kinetic friction force ($$f_k$$) acting on the sliding block ($$M$$): $$f_k = \mu_k \cdot N = \mu_k \cdot Mg$$

$$f_k = 0.02 \times 40 \times 10 = 8\ \text{N}$$

For the hanging block ($$m$$) moving downward:   $$mg - T = ma \quad \text{--- (1)}$$

For the sliding block ($$M$$) moving horizontally:   $$T - f_k = Ma \quad \text{--- (2)}$$

Adding equations (1) and (2) to eliminate tension ($$T$$):   $$mg - f_k = (m + M)a$$

$$(4 \times 10) - 8 = (4 + 40)a \implies 40 - 8 = 44a$$ 

$$a = \frac{32}{44} = \frac{8}{11}\ \text{m s}^{-2}$$