Question

Two blocks of masses $4kg$ and $10kg$ are connected by a flexible inextensible string as shown in given figure. The pulley is assumed to be light and frictionless. The coefficient of friction between the $4kg$ block and the suface is $0.2$ (${\mu }_s={\mu }_k=0.2$). The angle of inclination of the plane is ${30}^{\circ }$. Assuming the sytem to be released from rest, the acceleration of $10kg$ block is $(g=10m{s}^{-2})$

• A. $5.2m{s}^{-2}$
• B. $6.6m{s}^{-2}$
• C. $41.8m{s}^{-2}$
• D. $13m{s}^{-2}$

Answer 1

The correct option is A $5.2m{s}^{-2}$

In the scenario presented in the question, $4kg$ block slides up on the surface of inclined plane. So, friction acting will be kinetic in nature and acts downwards along the incline i.e
$f=f_k={\mu }_{k}N\cos \theta =\left(\right(0.2)4g\cos {30}^{\circ }$


acceleration of system:
$a=\frac{F_{net}}{m}$
solving for which get following:
$a=\left(\frac{10g-4gsin{30}^{\circ }-\left(0.2\right)\left(4\right)gcos{30}^{\circ }}{10+4}\right)$
$\Rightarrow a=\frac{(10-2-0.7)}{14}g$
$\Rightarrow a=\left(\frac{7.3}{14}\right)10$
$\Rightarrow a=5.2m{s}^{-2}$

Hence, the nearest correct answer is (A).