Question

Consider a system of two masses and a pulley shown in the figure. The coefficient of friction between the two blocks and also between block and table is $0.1 .$ Find the force $F$, that must be given to the $0.8\, kg$ block such that it attains acceleration of $5\, m / s ^{2}$. (Assume, acceleration due to gravity, $\left.g=10\, m / s ^{2} .\right)$

• A. 6.4 N
• B. 7.1 N
• C. 6.0 N
• D. 7.8 N

Answer 1

Answer: (A)

Given, acceleration of block, $a=5 \,m / s ^{2}$ and coefficient of friction between two blocks and table, $\mu=0.1$

Free body diagram for $0.8\, kg$ block.

For $0.8 \,kg$ block, friction force due to $0.2\, kg$ block is in opposite direction of applied force $F$.
$\therefore $ Equation of motion for $0.8 \,kg$ block,
$F-T-\mu R-\mu R_{1}=0.8 a$
putting the given values, we get
$F-T-0.1(0.8+0.2) g-0.1 \times 0.2 g=0.8 a$
$F-T-0.1 \times 10-0.1 \times 0.2 \times 10=0.8 \times 5 $
$ F-T=5.2\,\,\,...(i)$
Free body diagram for $0.2\, kg$ body,

For $0.2 \,kg$ block, friction force due to $0.8 \,kg$ block is in opposite direction of tension force.
So, $T-\mu R_{1}=0.2 a $
$T-0.1 \times 0.2 g=0.2 \times 5 $
$T-0.02 \times 10=1.0 $
$T=1.2\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$F-1.2=5.2$
$F=5.2+1.2=6.4 \,N$