Question

Two blocks of masses $5\text{kg}$ and $3\text{kg}$ are placed in contact over an inclined surface of angle ${37}^{\circ }$, as shown. ${\mu }_1$ is friction coefficient between $5\text{kg}$ block and the surface of the incline and similarly, ${\mu }_2$ is friction coefficient between the $3\text{kg}$ block and the surface of the incline. After the release of the blocks from the inclined surface, ($g=10{\text{m/s}}^2$)

• A. if ${\mu }_1=0.5$ and ${\mu }_2=0.3$, then $5\text{kg}$ block exerts $3\text{N}$ force on the $3\text{kg}$ block
• B. if ${\mu }_1=0.5$ and ${\mu }_2=0.3$, then $5\text{kg}$ block exerts $8\text{N}$ force on the $3\text{kg}$ block
• C. if ${\mu }_1=0.3$ and ${\mu }_2=0.5$, then $5\text{kg}$ block exerts $1\text{N}$ force on the $3\text{kg}$ block
• D. if ${\mu }_1=0.3$ and ${\mu }_2=0.5$, then $5\text{kg}$ block exerts no force on the $3\text{kg}$ block

Answer 1

Answer: (A), (D)

if ${\mu }_1=0.3$ and ${\mu }_2=0.5$, then $5\text{kg}$ block exerts no force on the $3\text{kg}$ block

Case-I: ${\mu }_1=0.5,{\mu }_2=0.3$
Along the incline, acceleration of $5\text{kg}$ block will be less than acceleration of $3\text{kg}$ block provided they move separately on the incline. The reason is greater friction coefficient of $5\text{g}$ block, as acceleration along the incline is
$g\sin \theta -\mu g\cos \theta$
So, both blocks will move together. In this case FBDs of both are shown.
For $5\text{kg}$ block
$m_{1}g\sin \theta +N-{\mu }_1{m}_{1}g\cos \theta =m_{1}a$
For $3\text{kg}$ block
$m_{2}g\sin \theta -N-{\mu }_2{m}_{2}g\cos \theta =m_{2}a$
Solving the two equations, we get $N=3\text{N}$
In second case, acceleration of $5\text{kg}$ block is more than acceleration of $3\text{kg}$. Hence, no normal force will act between them.