Question

Find the acceleration of block $A$ and $B$? Take $g=10m/s^2$

• A. $a_A=1{\text{m/s}}^2,a_B=10{\text{m/s}}^2$
• B. $a_A=2{\text{m/s}}^2,a_B=2{\text{m/s}}^2$
• C. $a_A=10{\text{m/s}}^2,a_B=2{\text{m/s}}^2$
• D. $a_A=0{\text{m/s}}^2,a_B=10{\text{m/s}}^2$

Answer 1

The correct option is D $a_A=0{\text{m/s}}^2,a_B=10{\text{m/s}}^2$

The FBDs for both block $A$ and $B$ are as shown
Given, external force $F=200N$
Fron the FBD we have
$f_{1\text{max}}=\mu N=0.5\left(5g\right)=25\text{N}$
$f_{2\text{max}}=\mu N=0.5\left(15g\right)=75\text{N}$

First we will check they will slide together (or) not?

We have
$f_{2\text{max}}$ available is $=\mu N=0.5\left(15g\right)=75\text{N}$
$f_{\text{ext}}\text{(Driving force)}>f_{2\text{max}}\left(\text{available friction}\right)$
By writing equation of force we have
$200-75=(10+5)a$
$125=15a$
$a=\frac{25}{3}{\text{m/s}}^2$

Now, we have to check for block $A$, to accelerate with this common acceleration, the minimum required driving force is
$F_{\text{required}}=ma$
$F_{\text{required}}=5\times \frac{25}{3}=41.6\text{N}$
Force required > available driving force $\left(f_{1\text{max}}\right)$
$\therefore a_A=0{\text{m/s}}^2$
So, they cannot slide together, but $B$ will slide individually.
Hence, acceleration of Block $B$ is given by

$200-75-25=10a$
$a_B=10{\text{m/s}}^2$