Two blocks of masses 2 kg and 4 kg are attached by an inextensible light string as shown in the figure. If a force of 120 N pulls the blocks vertically upward, the tension in the string is (take \(g = 10 \text{ ms}^{-2}\))
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For connected masses, always find the acceleration of the whole system first. Then, look at the block that has fewer forces acting on it (usually the bottom one) to solve for tension quickly.
Concept:
This problem is based on Newton's Second Law of Motion applied to a connected system of masses.
• System Acceleration: When a force acts on a system of connected masses, they accelerate together.
• Free Body Diagram (FBD): To find internal forces like tension, we must analyze the forces acting on an individual mass.
• Newton's Second Law: \(F_{net} = ma\).
Step 1: Calculate the common acceleration of the system.
Consider both blocks as a single system of mass \(M = 4 \text{ kg} + 2 \text{ kg} = 6 \text{ kg}\).
The upward force is \(F = 120 \text{ N}\). The downward force is the total weight \(W = Mg = 6 \times 10 = 60 \text{ N}\).
\[ F_{net} = F - Mg = Ma \]
\[ 120 - 60 = 6 \times a \]
\[ 60 = 6a \implies a = 10 \text{ ms}^{-2} \]
Step 2: Calculate the tension in the string.
To find the tension (\(T\)), analyze the lower block (2 kg mass) individually.
The forces acting on the 2 kg block are the tension \(T\) acting upward and its weight \(w = mg = 2 \times 10 = 20 \text{ N}\) acting downward.
\[ T - mg = ma \]
\[ T - 20 = 2 \times 10 \]
\[ T - 20 = 20 \implies T = 40 \text{ N} \]
