You drag a block with constant speedacross a table with friction. Explain in detail what you have to do in order to change to a constant speed of $\mathbf{\text{2v}}$on the same surface. (That is, the puzzle is to explain how it is possible to drag a block with sliding friction at different constant speeds.)

Equilibrium is the state of a system in which neither its state of motion nor its internal energy state changes over time, according to physics.A simple mechanical body is considered to be in equilibrium if it experiences neither linear nor angular acceleration; unless disturbed by an outside force, it will remain in that state indefinitely. Equilibrium occurs when the vector sum of all forces acting on a single particle equals zero.

It is given that$V$is the constant speed of the block across the tableis the coefficient of friction between the table and blockis the mass of the block.

In the picture,${\text{F}}_{\text{y}}$is present which is the normal reaction force. Friction force between the block and table is${\text{F}}_{\text{fric}}$ . The weight of the block is$\text{w = mg}$and the equilibrium position is,${\text{F}}_{\text{y}}\text{= w}$.

The normal reaction is proportional to the frictional force, write the expression.

role="math" localid="1668490557032" ${\text{F}}_{\text{fric}}{\text{αμF}}_y$

Remove the proportionality constant.

role="math" localid="1668490625164" ${\text{F}}_{\text{fric}}{\text{=μF}}_y$ …….(1)

The frictional force is the horizontal force and$F$is the pulling force acting on the block.

Write the net force acting on the block from the diagram.

${\text{F -F}}_{\text{fric}}\text{= ma}$

From the given information, the acceleration is0. Let the speed, $v$ of the block is constant.

Substitute$a=0$into the equation of net force acting on the block.

$$\begin{gathered} {\text{F -F}}_{\text{fric}}\text{= m}\left(\text{0}\right) \\ {\text{F -F}}_{\text{fric}}\text{= 0} \\ {\text{F =F}}_{\text{fric}} \end{gathered}$$

From equation (1) it is concluded that the frictional force is depending of the weight of the block and is independent of the speed of the block.

When the speed of the block is doubled that is $2v$, the acceleration will remain constant that is 0.

Substitute $a=0$ into the equation of net force acting on the block.

$$\begin{gathered} {\text{F -F}}_{\text{fric}}\text{= m}\left(\text{0}\right) \\ {\text{F -F}}_{\text{fric}}\text{= 0} \\ {\text{F =F}}_{\text{fric}} \end{gathered}$$

Therefore, the frictional force of a moving block at a constant speed on a surface with friction remains constant, but it does not depend on the speed of the block resulting for the pulling force to be unchanged.