Question

choose correct answer with neat explanation​

Answer 1

The statements $m_{1} = \mu_{s}$ and $m_{2} = \mu_{k}$ make no sense.

Answer: A, B, C.

Explanation:

1. Assuming the system does not move:

The friction acting between $m_{2}$ and the floor is static (since the block is assumed to not be moving.

Now, since the block $m_{1}$ is also not moving, and tension is the only other force to balance out the pull of gravity ($m_{1}g$), the tension must be equal to $m_{1}g$.

The same tension is applied on the block $m_{2}$ (assuming the string is massless), and hence, for that block to stay in equilibrium, the static friction limit must be greater than or equal to the tension.

The static friction limit is: $\mu_{s}m_{2}g$, since the normal is $m_{2}g$.

So: $\mu_{s}m_{2}g >= m_{1}g$

Or, $\mu_{s}m_{2} >= m_{1}$

In this case, as we've already deduced, the tension in the string would be $m_{1}g$, and the static friction too, would be $m_{1}g$.  A, B

2. Assuming the system moves:

From case 1, we have deduced the condition:

$\mu_{s}m_{2} >= m_{1}$

If the above condition is disobeyed, the system will begin moving. However, a change would occur: that the friction between the block $m_{2}$ and the floor becomes kinetic in nature, and shall use the coefficient $\mu_{k}$.

The magnitude of the kinetic friction is always $\mu_{k}m_{2}g$.  C