Two blocks A and B are connected to each other by a string and a spring; the string passes over a fr

Q: Why is there no friction between block $A$ and block $C$?

Friction requires a normal reaction force ($f = \mu N$). Since block $A$ is hanging vertically and no horizontal force is pushing it into block $C$, the normal reaction is zero. Hence, friction is zero.

Q: Why do we take acceleration as zero?

The problem explicitly states that both blocks move with "uniform speed". Uniform speed means velocity is constant, which by definition means acceleration is exactly zero.

Q: Is the tension in the string equal to the spring force?

Yes! The string and the spring are connected in series. Since the string is massless, it simply transmits the force exerted by the spring without any loss. Therefore, $T = kx$.

Q: Why don't we use $g = 10 \text{ m/s}^2$?

Unless specified otherwise, standard practice in JEE is to use $g = 9.8 \text{ m/s}^2$. Using $10$ here would give $x = 0.0102 \text{ m}$, which leads to a slightly different final answer. The exact integer answer $10 \text{ kg}$ confirms $9.8$ was intended.

Q: What if the blocks were accelerating?

If the blocks were accelerating, we would use Newton's Second Law ($F_{\text{net}} = ma$). For block $A$, it would be $m_A g - kx = m_A a$, and for block $B$, $T - \mu m_B g = m_B a$. The problem would be much more complex!

Question

Two blocks $A$ and $B$ are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown in the figure. Block $B$ slides over the horizontal top surface of a stationary block $C$ and the block $A$ slides along the vertical side of $C$, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is $0.2$. Force constant of the spring is $1960	ext{ N/m}$. If mass of block $A$ is $2	ext{ kg}$. Calculate the mass of block $B$ and the energy stored in the spring.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing right in front of this mechanical system. We have block $B$ sliding smoothly over the top of block $C$, while block $A$ is descending along the vertical face.

The most critical piece of information given in the problem is that both blocks are moving with a **uniform speed**. In the realm of physics, uniform speed is a magical phrase. It instantly tells us that the acceleration of the entire system is exactly zero.

According to Newton's Second Law, if the acceleration is zero, the net force acting on every single component of the system must also be zero. This means the system is in a state of dynamic equilibrium, and we can simply balance the opposing forces.

### The Illusion of Friction on Block A

Let's focus our attention on block $A$. It is sliding down the vertical wall of block $C$. A common trap here is to assume there is some friction acting upwards against its motion.

However, friction requires two surfaces to be pressed against each other. Is there any horizontal force pushing block $A$ into block $C$? No. The string pulls it straight up, and gravity pulls it straight down.

Because there is no horizontal pressing force, the normal reaction $N_{AC}$ between block $A$ and block $C$ is strictly zero. Consequently, the frictional force is also zero. Block $A$ is essentially in free fall, restrained only by the spring above it.

### The Master Equations

Now, let's draw the Free Body Diagram for block $A$. Since it is in equilibrium, the upward forces must perfectly balance the downward forces. The downward force is its weight, $m_A g$. The upward force is the restoring force of the stretched spring, $F_s = kx$.

Equating these, we get our first master equation:

$$kx = m_A g$$

Next, we shift our focus to block $B$. It is moving horizontally, so its forward and backward forces must cancel out. The forward force is the tension $T$ from the string. The backward force is the kinetic friction from the surface of block $C$, which is given by $f_k = \mu m_B g$.

This gives us our second master equation:

$$T = \mu m_B g$$

### The Beautiful Connection

Here is where the physics gets truly elegant. The string pulling block $B$ is directly connected to the spring holding block $A$. Because the string is ideal and massless, the tension $T$ everywhere in the string must be exactly equal to the force exerted by the spring.

Therefore, we can confidently state that $T = kx$. And from our first equation, we already know that $kx = m_A g$. This creates a beautiful chain of equality:

$$T = kx = m_A g$$

### Final Calculation

We can now substitute $m_A g$ for the tension $T$ in our friction equation for block $B$. This yields:

$$\mu m_B g = m_A g$$

Notice how the acceleration due to gravity, $g$, elegantly cancels out from both sides. Rearranging the equation to solve for the mass of block $B$, we get:

$$m_B = \frac{m_A}{\mu}$$

Substituting the given values ($m_A = 2	ext{ kg}$ and $\mu = 0.2$), we find:

$$m_B = \frac{2}{0.2} = 10	ext{ kg}$$

Finally, we need to calculate the energy stored in the stretched spring. First, we find the extension $x$ using our first equation:

$$x = \frac{m_A g}{k} = \frac{2 	imes 9.8}{1960} = 0.01	ext{ m}$$

The elastic potential energy $U$ stored in the spring is given by the formula:

$$U = \frac{1}{2} kx^2$$

Plugging in our values, we get:

$$U = \frac{1}{2} (1960) (0.01)^2 = 980 	imes 0.0001 = 0.098	ext{ J}$$

And there we have it! By carefully breaking down the forces and recognizing the state of dynamic equilibrium, we have flawlessly unraveled this pulley-spring system.

Visualized Solution

Visualizing the Setup

Normal Reaction on Block $A$

Free Body Diagram of Block $A$

Calculating Spring Extension $(x)$

Free Body Diagram of Block $B$

Relating Tension and Spring Force

Calculating Mass of Block $B$

Energy Stored in the Spring

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Illusion of Friction on Block A

The Master Equations

The Beautiful Connection

Final Calculation

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