Animated Solution for Physics - Laws of Motion: Two masses $m_1=5\text{ kg}$ and $m_2=10\text{ kg}$ connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is $0.15$. The minimum weight $m$ that should be put on top of $m_2$ to stop the motion is

A Tug of War

Imagine a classic physics scenario: a block resting on a table, connected by a string over a pulley to another block hanging off the edge. The hanging block, driven by gravity, desperately wants to pull the entire system down. The block on the table, however, is gripping the rough surface, trying to hold its ground.

In our specific problem, the hanging mass $m_1$ is $5\text{ kg}$, and the block on the table $m_2$ is $10\text{ kg}$. The surface isn't perfectly smooth; it has a coefficient of friction $\mu =0.15$. The question asks: how much extra mass $m$ do we need to pile on top of $m_2$ to completely stop the system from moving?

To solve this, we need to understand the forces battling each other.

The driving force is the weight of the hanging mass. This force is transmitted through the tension in the string and pulls the blocks on the table horizontally.

Driving Force = $m_{1}g$

The opposing force is the static friction between the bottom block $m_2$ and the table. Friction depends on two things: the roughness of the surfaces (the coefficient $\mu$) and how hard the surfaces are pressed together (the normal force $N$).

Because we are adding an extra mass $m$ on top of $m_2$, the total mass pressing down on the table is $(m_2+m)$. Therefore, the normal force is $(m_2+m)g$.

Maximum Static Friction = $\mu N=\mu (m_2+m)g$

For the system to remain stationary, the opposing friction must be strong enough to withstand the driving pull. Mathematically, this means the pulling force must be less than or equal to the maximum possible static friction:

Notice how the acceleration due to gravity, $g$, appears on both sides of the inequality. This is a beautiful moment in physics—it tells us that this balancing act would work exactly the same way on the Moon or on Mars! We can safely cancel $g$ from both sides:

Now, let's plug in the numbers we know: - $m_1=5\text{ kg}$ - $m_2=10\text{ kg}$ - $\mu =0.15$

Substituting these into our simplified inequality:

Let's expand the right side:

Subtract $1.5$ from both sides to isolate the term with $m$:

Finally, divide by $0.15$:

Our math tells us that we need at least $23.33\text{ kg}$ of extra mass to keep the system from sliding.

Let's look at the given options: (a) $18.3\text{ kg}$ (b) $27.3\text{ kg}$ (c) $43.3\text{ kg}$ (d) $10.3\text{ kg}$

If we choose $18.3\text{ kg}$ or $10.3\text{ kg}$, the mass is less than our required $23.33\text{ kg}$, and the system will accelerate downwards. Both $27.3\text{ kg}$ and $43.3\text{ kg}$ will successfully stop the motion. However, the question asks for the minimum weight from the given choices that will do the job. The smallest valid option is $27.3\text{ kg}$.

Therefore, the correct answer is 27.3 kg.