Physics: Pulley Weight Speed Calculation

Hey physics fans! Today we're diving into a classic problem involving pulleys and weights. You know, those times when you're trying to figure out how fast things are moving when they're connected by a string over a wheel? Well, this one's a doozy, but don't sweat it, guys, we'll break it down step-by-step. The question is: On a cord stretched over a fixed block, weights of 200g and 300g are suspended. What will be the speed of the weights 2 seconds after the start of motion? This is a fantastic way to understand Newton's laws in action, specifically the second law (F=ma), and how it applies to systems with multiple masses. We'll need to get a bit into the nitty-gritty of forces, acceleration, and tension to solve this one, but trust me, by the end, you'll have a solid grasp on how to tackle these types of pulley problems. So, grab your calculators, maybe a notepad, and let's get this physics party started!

Understanding the Setup: The Atwoord Machine

Alright, let's get real about what we're dealing with here. We have a simple yet super effective setup called an Atwood machine. Think of it as two masses connected by a string, and that string is draped over a pulley. This pulley is fixed, meaning it doesn't move up or down – it just rotates. Now, we've got two weights: one is 200 grams, and the other is a bit heavier at 300 grams. Because there's a difference in mass, the heavier weight will start to pull the lighter weight up, and the lighter weight will pull the heavier one down. This creates a system where both masses move together with the same acceleration. This is a key point, guys! They are yoked together by the string, so whatever speed and acceleration one experiences, the other does too, just in the opposite direction. The total mass of the system is important, but more crucial for calculating the acceleration is the difference in mass. This difference is what drives the motion. We're also assuming an ideal scenario here – the string is massless and inextensible (it doesn't stretch), and the pulley is massless and frictionless. In the real world, these things would introduce complexities, but for learning the fundamental physics, these assumptions are super helpful. So, with our 200g and 300g masses, we know motion is inevitable. The bigger mass is going to win the tug-of-war and pull the system into motion. Our goal is to find out how fast these masses are moving after a specific time, 2 seconds, which means we need to figure out their acceleration first.

The Forces at Play: Newton's Second Law is Your Best Friend

Now, let's talk forces, because this is where the magic happens. For each mass, we have two main forces acting on it: gravity pulling it down and tension in the string pulling it up. Newton's Second Law of Motion, F = ma (Force equals mass times acceleration), is our golden ticket here. We need to apply this law to each mass individually.

Let's label the masses: $m_1 = 200g = 0.2kg$ (the lighter mass) and $m_2 = 300g = 0.3kg$ (the heavier mass). We'll convert grams to kilograms because, in physics, we usually work with standard SI units, and acceleration will come out in meters per second squared ($m/s^2$). The acceleration for both masses will be the same, let's call it '$a$'.

For the heavier mass ($m_2$), gravity pulls it down with a force $F_{g2} = m_2 imes g$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). The tension in the string, $T$, pulls it upwards. Since $m_2$ is accelerating downwards, the net force on $m_2$ is $F_{net2} = F_{g2} - T$. According to Newton's Second Law, this net force equals $m_2 imes a$. So, our first equation is:

$m_2 imes g - T = m_2 imes a$ (Equation 1)

Now, let's look at the lighter mass ($m_1$). Gravity pulls it down with a force $F_{g1} = m_1 imes g$. The tension $T$ pulls it upwards. Since $m_1$ is accelerating upwards, the net force on $m_1$ is $F_{net1} = T - F_{g1}$. According to Newton's Second Law, this net force equals $m_1 imes a$. So, our second equation is:

$T - m_1 imes g = m_1 imes a$ (Equation 2)

See how the tension $T$ is the same for both? That's because it's the same string. Also, notice the direction of the net force and acceleration are opposite for the two masses. This is the core of setting up these problems. We have two equations and two unknowns ($T$ and $a$), which means we can solve for them!

Solving for Acceleration: Unraveling the Mystery

Alright, guys, we've got our two equations, and our mission now is to find the acceleration, '$a$'. The easiest way to do this is to eliminate the tension '$T$' from the equations. We can do this by adding Equation 1 and Equation 2 together. Check this out:

$(m_2 imes g - T) + (T - m_1 imes g) = (m_2 imes a) + (m_1 imes a)$

Notice how the '$T$' terms cancel each other out? That's super convenient! We're left with:

$m_2 imes g - m_1 imes g = m_2 imes a + m_1 imes a$

We can factor out '$g$' on the left side and '$a$' on the right side:

$g imes (m_2 - m_1) = a imes (m_1 + m_2)$

Now, all we have to do is rearrange this to solve for '$a$'. We just divide both sides by $(m_1 + m_2)$:

a = g imes rac{(m_2 - m_1)}{(m_1 + m_2)}

This is the general formula for the acceleration of an Atwood machine with two masses and a frictionless, massless pulley! Pretty neat, right? Now we can plug in our values.

We have $m_1 = 0.2kg$, $m_2 = 0.3kg$, and g oldsymbol{acksimeq} 9.8 m/s^2.

a = 9.8 m/s^2 imes rac{(0.3kg - 0.2kg)}{(0.2kg + 0.3kg)}

a = 9.8 m/s^2 imes rac{0.1kg}{0.5kg}

$a = 9.8 m/s^2 imes 0.2$

$a = 1.96 m/s^2$

So, the acceleration of the system is $1.96 m/s^2$. This means that for every second the system is in motion, the speed of the weights increases by $1.96 m/s$. Pretty cool, huh?

Calculating the Final Speed: Putting It All Together

We're almost there, folks! We’ve found the acceleration of the system, which is $1.96 m/s^2$. The question asks for the speed of the weights after 2 seconds. We can use another fundamental kinematic equation for this. Since the weights start from rest (implied by "after the start of motion"), their initial velocity ($v_0$) is $0 m/s$. The equation we need is:

$v = v_0 + a imes t$

Where:

• '$v$' is the final velocity (what we want to find).
• '$v_0$' is the initial velocity ($0 m/s$).
• '$a$' is the acceleration ($1.96 m/s^2$).
• '$t$' is the time ($2 s$).

Let's plug in the numbers:

$v = 0 m/s + (1.96 m/s^2) imes (2 s)$

$v = 0 m/s + 3.92 m/s$

$v = 3.92 m/s$

And there you have it! The speed of the weights after 2 seconds will be 3.92 m/s. The heavier mass will be moving downwards at $3.92 m/s$, and the lighter mass will be moving upwards at the same speed.

Why This Matters: Real-World Applications of Pulley Systems

So, why do we even bother with these physics problems, right? Well, understanding pulley systems like the Atwood machine is super important because pulleys are everywhere! Think about it – they are used in cranes to lift heavy loads, in elevators to move people up and down, in gym equipment to provide resistance, and even in simple things like raising a flag on a flagpole. The principles we used, like Newton's laws and kinematic equations, are the bedrock of mechanical engineering and everyday physics. By solving this problem, you've not only tackled a specific physics question but also gained insight into how forces, motion, and acceleration work together in real-world mechanical systems. Imagine a construction worker using a pulley to lift concrete – they need to understand the forces involved to operate safely and efficiently. Or a rock climber using a pulley system for safety and ascent. Even the mechanics designing an elevator system rely on these fundamental calculations to ensure smooth and safe operation. Understanding the interplay between mass, gravity, tension, and acceleration allows us to predict and control motion. This basic Atwood machine problem, though seemingly simple, is a stepping stone to understanding more complex machines and dynamic systems. So next time you see a pulley, you'll know there's some cool physics at play!

Key Takeaways and Further Exploration

To wrap things up, guys, we've successfully calculated the speed of the weights after 2 seconds in our Atwood machine problem. The key steps involved were: identifying the forces acting on each mass, applying Newton's Second Law to set up equations for each mass, solving those equations simultaneously to find the acceleration of the system, and finally, using a kinematic equation to find the final velocity. The acceleration we found was $1.96 m/s^2$, and the final speed after 2 seconds was $3.92 m/s$. Remember, these calculations assume ideal conditions (massless string and pulley, no friction). In a real-world scenario, you'd have to account for these factors, which would slightly alter the results. If you want to explore further, try changing the masses and see how the acceleration and final speed change. What happens if the masses are equal? What if one mass is significantly larger than the other? You could also investigate the effect of a non-ideal pulley (e.g., with friction or mass) or a stretching string. These extensions can lead to more complex differential equations but deepen your understanding of mechanical systems. Keep practicing, keep questioning, and you'll master this physics stuff in no time! Keep up the great work, future physicists!