Question

Coefficient of friction between the two blocks is $0.3$, whereas the surface $AB$ is smooth. Choose the correct options ($g=10{\text{m/s}}^2$)

• A. Acceleration of the system of masses is $\frac{88}{15}{\text{m/s}}^2$.
• B. Net force acting on $3\text{kg}$ mass is greater than that on $2\text{kg}$ mass.
• C. Tension $T_2>T_1$
• D. Since $10\text{kg}$ mass is accelerating downwards, so net force acting on it should be greater than any of the two blocks shown in figure.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Acceleration of the system of masses is $\frac{88}{15}{\text{m/s}}^2$.
B Net force acting on $3\text{kg}$ mass is greater than that on $2\text{kg}$ mass.
C Tension $T_2>T_1$

(a) Let the acceleration of each block be $a$.
$10g-T_2=10a,T_2-T_1-f=3a$
$T_1-f=2a$, where $f=0.3\times 2g=6\text{N}$
From the above equations:
$10g-2f=15a\Rightarrow 10\times 10-2\times 6=15a$
$\Rightarrow a=\frac{88}{15}{\text{m/s}}^2$
$T_2=10g-10a=10\times 10-10\times \frac{88}{15}=41.3\text{N}$
$T_1=f+2a=6+2\times \frac{88}{15}=17.7\text{N}$
Clearly $T_2>T_1$
(b) This is correct because of the greater mass of $3\text{kg}$, since acceleration is same for both.
(d) This is incorrect because the net force acting on $10\text{kg}$ mass is greater due to its larger mass, not due to its acceleration downward.