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Q5.15(i):
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $F = 600 N$ is applied to (i) A, along the direction of string. What is the tension in the string in each case?

Solution :

Given Variables & Initial Setup

Let $m_1 = 10\text{ kg}$ (Body A) and $m_2 = 20\text{ kg}$ (Body B). Both bodies are connected by a light (massless) string on a smooth (frictionless) horizontal surface. An external force $F = 600\text{ N}$ is applied to body A (the 10 kg mass) in the direction of the string, pulling body B behind it.

10kg 20kg Tension T F = 600N

Step 1: Calculating the Acceleration of the System

Since the bodies are tied together by a light string, they move as a single composite system with common acceleration $a$ [Per Newton's Second Law of Motion].

The total mass of the system is $M = m_1 + m_2 = 10\text{ kg} + 20\text{ kg} = 30\text{ kg}$.

Applying Newton's Second Law to the composite system:

$F = M \cdot a$ $600\text{ N} = 30\text{ kg} \cdot a$ $a = \frac{600}{30} = 20\text{ m/s}^2$

Step 2: Free Body Diagram Analysis for Tension ($T$)

To find the tension $T$ in the string, we isolate body B ($m_2 = 20\text{ kg}$). The only horizontal force acting on body B is the tension $T$ exerted by the string [Per Newton's Third Law of Motion].

Applying Newton's Second Law to body B:

$T = m_2 \cdot a$

Step 3: Solving for Tension

Substituting the known values into the equation derived in Step 2:

$T = 20\text{ kg} \times 20\text{ m/s}^2$ $T = 400\text{ N}$

Summary of Dynamics

Parameter Value
Total Mass ($M$) $30\text{ kg}$
System Acceleration ($a$) $20\text{ m/s}^2$
Tension ($T$) $400\text{ N}$

Final Solution: The tension in the string when the force is applied to the 10 kg mass is 400 N.


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