Question 42 of 210

A block of mass $5kg$ is (i) pushed in case $\left(A\right)$ and (ii) pulled in case $\left(B\right),$ by a force $F=20N,$ making an angle of ${30}^{o}$ with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is $\mu =0.2.$ The difference between the accelerations of the block, in case $\left(B\right)$ and case $\left(A\right)$ will be:
$\left(g=10 m{s}^{-2}\right)$

A$3.2m{s}^{-2}$B$0m{s}^{-2}$C$0.8m{s}^{-2}$D$0.4m{s}^{-2}$

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Answer: (C) $0.8m{s}^{-2}$

For Block A:

$N=mg+F\sin ⁡30^{\circ}$
$=5\times 10+20\times \dfrac{1}{2}$
$=60N$
${f}_{r}=\left(0.2\right)\left(60\right)=12N$
$F\cos ⁡30^{\circ}–{f}_{r}=ma$
$20\dfrac{\sqrt{3}}{2}$ $-12=5\times a$
$\Rightarrow {a}_{A}=\dfrac{10\sqrt{3}-12}{5}$ ……(i)
For Block B:

$F\sin ⁡30^{\circ}+N=mg$
$N=5\times 10–20\times \left(\dfrac{1}{2}\right)$
$=40N$
${f}_{r}=0.2\times 40=8N$
$Fcos30^{\circ}–{f}_{r}=ma$
$20\left(\dfrac{\sqrt{3}}{2}\right)-8=5\times a$
$\Rightarrow {a}_{B}=\dfrac{10\sqrt{3}-8}{5}$
${a}_{B}-{a}_{A}=\dfrac{4}{5}=0.8m/{s}^{2}$