Question

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

• A. $0.4{\text{ms}}^{-2}$
• B. $3.2{\text{ms}}^{-2}$
• C. $0.8{\text{ms}}^{-2}$
• D. $0{\text{ms}}^{-2}$

Answer 1

The correct option is C $0.8{\text{ms}}^{-2}$

Case $\text{I}:$ Block is pushed over surface


Free body diagram of block is

In this case, normal reaction,
$N=mg+F\sin {30}^{\circ }=(5\times 10)+(20\times \frac{1}{2})=60\text{N}$

$[\text{Given},m=5\text{kg},F=20\text{N}]$

Force of friction, $f=\mu N=0.2\times 60=12\text{N}[\because \mu =0.2]$

So, net force causing acceleration $\left(a_1\right)$ is

$F_{\text{net}}=m{a}_1=F\cos {30}^{\circ }-f$

$\Rightarrow m{a}_1=20\times \frac{\sqrt{3}}{2}-12$

$\therefore a_1=\frac{10\sqrt{3}-12}{5}=1{\text{ms}}^{-2}$

Case $\text{II}:$ Block is pulled over the surface


Net force causing acceleration is

$F_{\text{net}}=F\cos {30}^{\circ }-f=F\cos {30}^{\circ }-\mu N$

$\Rightarrow F_{\text{net}}=F\cos {30}^{\circ }-\mu (mg-F\sin {30}^{\circ })$

If acceleration is now $a_2$, then

$a_2=\frac{F_{\text{net}}}{m}=\frac{F\cos {30}^{\circ }-\mu (mg-F\sin {30}^{\circ })}{m}$

$=\frac{20\times \frac{\sqrt{3}}{2}-0.2(5\times 10-20\times \frac{1}{2})}{5}=\frac{10\sqrt{3}-8}{5}$

$\Rightarrow a_2=1.8{\text{ms}}^{-2}$

So, difference between $a_1\&a_2$ is,

$\Delta a=a_2-a_1=1.8-1=0.8{\text{ms}}^{-2}$

Hence, option $\left(\text{C}\right)$ is correct.