Question

A mass of $2\text{kg}$ is placed on a trolley of $20\text{kg}$ sliding on a smooth surface. The coefficient of friction between the mass and the surface of the trolley is $0.25$. A horizontal force of $2\text{N}$ is applied to the mass. The acceleration of the system and the frictional force between the mass and surface of trolley are (Take $g=10m/s^2$)

• A. $1.8{\text{ms}}^{-2},0.09\text{N}$
• B. $0.9{\text{ms}}^{-2},18\text{N}$
• C. $0.09{\text{ms}}^{-2},1.8\text{N}$
• D. $1{\text{ms}}^{-2},2\text{N}$

Answer 1

The correct option is C $0.09{\text{ms}}^{-2},1.8\text{N}$

The FBDs of the system is shown as


From the FBD of block,
$N_1=2g=20\text{N}$
From the FBD of trolley,
$N_2=N_1+20g=20+200=220\text{N}$

Limiting frictional force $f_{max}=\mu N_1=0.25\times 20=5\text{N}$
$\because f_{max}>$ applied external force of $2N$, the block and trolley will not have relative motion and they will move together.

Let the common acceleration be $a_c$
Then, $2=(20+2)a_c$
$\Rightarrow a_c=\frac{2}{22}=\frac{1}{11}=0.09{\text{ms}}^{-2}$

Frictional force is given from the FBD of trolley as
$f=20a_c=20\times 0.09=1.8\text{N}$