Question

A block $B$ of mass $5kg$ is placed on a slab $A$ of mass $20kg$ which lies on a frictionless surface as shown in the figure. The coefficient of static friction between the block and the slab is $0.4$ and that of kinetic friction is $0.2$. If a force $F=25N$ acts on $B$, the acceleration of the slab will be $(g=10m{s}^{-2})$.

• A. $0.4m{s}^{-2}$
• B. $0.5m{s}^{-2}$
• C. $1m{s}^{-2}$
• D. Zero

Answer 1

The correct option is B $0.5m{s}^{-2}$

Maximum magnitude of static friction force$=0.4\times 5\times 10m/s^2$

$=20N$

Now the force applied on the block is $25N$, while the static frictional force is $20N$, this implies that both the slab & box will move at different acceleration rates.

From Newton's second law,

$N_{sb}=m_{block}\times g$

$F-f_k=m_{block}\times g$

Also, $f_k=m_{slab}\times a_{slab}$

and $f_k={\mu }_k\times N_{sb}$

$\Rightarrow m_{slab}\times a_{slab}={\mu }_k\times N_{sb}$

$\Rightarrow m_{slab}\times a_{slab}={\mu }_k\times m_{block}\times g$

$\Rightarrow a_{slab}=\frac{{\mu }_k\times m_{block}\times g}{m_{slab}}$

$=\frac{0.2\times 5\times 10}{20}=0.5m/s^2$