Animated Solution for Physics - Work, Energy, and Power: A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36\text{ kg}$ and $0.72\text{ kg}$. Taking $g=10{\text{ ms}}^{-2}$, find the work done (in Joule) by string on the block of mass $0.36\text{ kg}$ during the first second after the system is released from rest.

A Symphony of Forces

Imagine a classic physics setup: a smooth, frictionless pulley with a light, inextensible string draped over it. Hanging from the ends of this string are two blocks of unequal masses. One block is a lightweight contender at $0.36\text{ kg}$, while the other is exactly twice as heavy at $0.72\text{ kg}$. When we release this system from rest, gravity takes over. The heavier block plunges downwards, pulling the lighter block upwards.

But we aren't just interested in watching them move. We have a specific mission: to calculate the exact amount of work done by the string on the lighter block during the very first second of its journey.

To unlock this, we need to remember the fundamental definition of work: $W=F\cdot s\cdot \cos \theta$. In our scenario, the force $F$ doing the work is the Tension $T$ in the string, and $s$ is the upward displacement of the block. To find both $T$ and $s$, we must first uncover the hidden engine driving this entire system—its acceleration.

Finding the Acceleration

Because the string is inextensible, both blocks are locked into a synchronized dance; they must move with the exact same magnitude of acceleration, $a$. Instead of getting bogged down by the internal tension forces right away, we can look at the two blocks as a single, unified system.

The net force driving this system is the difference in their weights. The heavier block pulls down with a force of $m_{2}g$, while the lighter block resists with a force of $m_{1}g$. According to Newton's Second Law, this net pulling force must equal the total mass of the system multiplied by its acceleration.

Let's plug in our specific masses:

Simplifying this fraction reveals a beautifully clean result: the acceleration of our system is exactly one-third of gravity, or $a=\frac{g}{3}$.

Tracking the Displacement

Now that we know how fast the system is speeding up, we can figure out exactly how far the lighter block travels in that crucial first second. Since the system is released from rest, its initial velocity $u$ is zero.

We can call upon the trusty second equation of kinematics:

Substituting our known values ($u=0$, $t=1\text{ s}$, and $a=\frac{g}{3}$), we get:

So, in one second, the lighter block is hoisted upwards by a distance of $\frac{g}{6}$ meters.

Uncovering the Tension

We have the displacement, but to calculate work, we need the force. It's time to zoom in and draw a Free Body Diagram exclusively for the lighter block, $m_1$.

Two forces are acting on it: the string pulling it up with Tension $T$, and gravity pulling it down with weight $m_{1}g$. Because we know the block is accelerating upwards, the upward Tension must be winning this tug-of-war. We can write its specific equation of motion:

Rearranging to solve for $T$, we see that the string must support the block's weight and provide the extra force needed to accelerate it:

Let's substitute our values:

The tension in the string is a constant $0.48g$ Newtons.

Work Done

We have finally gathered all the pieces of our puzzle. The Tension force is $T=0.48g$ acting upwards. The displacement is $s=\frac{g}{6}$ also directed upwards. Because the force and the displacement are perfectly aligned in the same direction, the angle $\theta$ between them is $0^{\circ }$, and $\cos (0^{\circ })=1$.

Let's calculate the work done:

The problem instructs us to take the acceleration due to gravity as $g=10{\text{ ms}}^{-2}$. Substituting this final piece:

And there we have it. In the first second of motion, the string performs exactly 8 Joules of work to lift the lighter block.