Calculate the acceleration of the block and trolley system shown in the figure. The coefficient of kinetic friction between the trolley and the surface is 0.05 (g = 10 m/s², mass of the string is negligible and no other friction exists).                                                         

The given situation is shown in the following diagram.

If a be the acceleration of the system, then equation of motion of 10 kg trolley,

T - $\mu$R = 10a

$\Rightarrow T-0.05\times 10g=10a (\text{ Given, }\mu =0.05,R=10g)$

$\Rightarrow T - 5 = 10a .......\left(\mathrm{i}\right)$

Equation of motion of 2kg block,

$2g - T = 2a$

$\Rightarrow 20 - T = 2a ......\left(\mathrm{ii}\right)$

Adding Eqs. (i) and (ii), we get

$15=12a\Rightarrow a=\frac{15}{12}=\frac{5}{4}=1.25{ \mathrm{ms}}^{-2}$