Question

All surfaces are smooth except that between $‘m^{\prime }$ and $‘2m^{\prime }$. The acceleration (in ${\text{m/s}}^2$) of the mass $‘m^{\prime }$ is

• A. $\frac{2g}{5}$
• B. $\frac{g}{5}$
• C. $\frac{3g}{5}$
• D. $\frac{4g}{5}$

Answer 1

The correct option is A $\frac{2g}{5}$

The FBDs of the blocks are


From the FBDs, we have
$N_1=mg$ and
$f_1=T=2mg$

Also, $f_{\text{1 max}}=\mu N_1=\mu mg=\frac{mg}{2}$

Assuming the system to move together with common acceleration $a$,
$a=\frac{\text{(Supporting Force - Opposing Force)}}{\text{Total mass}}$
$=\frac{2mg}{5m}=\frac{2g}{5}$
(because friction $f_1$ is an internal force for the system, it is not considered)

For mass ${}^{\prime }{m}^{\prime }$:
$f_1=ma=\frac{2mg}{5}$
$\because f_1<f_{\text{1 max}}$, our assumption is right.
Hence, they will move together with acceleration $a=\frac{2g}{5}$