Question

Two blocks of mass 5 kg and 10 kg are kept on a rough horizontal surface as shown in figure A force F is applied on upper block of mass 10 kg Now choose the correct statement(s) regarding the system

• A. If magnitude of applied force is 2N then frictional force between the blocks is also 2N
• B. The acceleration of 10 kg block is 2 $m/s^2$ when applied force is 30 N
• C. The minimum value of F for which 5 kg block begin to slide if coefficient of friction between blocks is changed to 0.5 is 45 N
• D. 5 kg block will never move on the ground for any value of F if coefficient of friction between blocks is 0.1 and between block and surface is 0.3

Answer 1

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

The correct options are
A If magnitude of applied force is 2N then frictional force between the blocks is also 2N
B The acceleration of 10 kg block is 2 $m/s^2$ when applied force is 30 N
C The minimum value of F for which 5 kg block begin to slide if coefficient of friction between blocks is changed to 0.5 is 45 N
D 5 kg block will never move on the ground for any value of F if coefficient of friction between blocks is 0.1 and between block and surface is 0.3

For 10kg block, f is frictional force

$f_{1max}={\mu }_1{\mu }_2={\mu }_1\left(10g\right)=0.1(10\times 10)=10N$

i) if $F=2N$ then acting friction

$f_1=F=2N$

(A) is correct.

for 5 kg block;

$f_{2max}=\mu (10+5)g=0.3\times 15\times 10=45N$

ii) if $F=30N$ then;

$F-F_{1max}=10a$

$10a=30-10=20$

$a=2m/s^2$

(B) is correct.

iii) if ${\mu }_1$ changed to 0.5;

then $f_{1max}={\mu }_1\left(10g\right)=0.5\times 10\times 10=50N$

In the case acting friction will be $F_{2max}$

Since,

$F_{1max}=F_{2max}$

Block 10kg and 5kg both move together then $F_{min}$ to begin is

$F_{min}-F_{max}=0$

$F_{min}=F_{2max}=45N$

(c) is correct.

iv) if $\mu =0.1$ , then $f_{max}=10N$ and,

$f_{2max}=45N$

$f_{2max}>f_{1max}$ always for any any value of $N$. Hence, 5kg block never move on the ground.

So, option (A), (B), (C), and (D) are correct.