Block and Trolley System NEET Solution | Acceleration & Tension Explained

NEET Physics Example 4.9: Block and Trolley System (Step-by-Step)

- Dr.Sanjaykumar Pawar 

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Physics NEET diagram showing a block and trolley system with forces, tension, friction, and acceleration clearly labeled.
       

Example 4.9 – Block and Trolley System (Easy NEET Notes)

Given:

• Mass of hanging block, \( m_1 = 3\,kg \)
• Mass of trolley, \( m_2 = 20\,kg \)
• Coefficient of kinetic friction, \( \mu_k = 0.04 \)
• Acceleration due to gravity, \( g = 10\,m\,s^{-2} \)
• String is light (massless) and inextensible
• Pulley is smooth (frictionless)

To Find:

1. Acceleration of the system \( a \)
2. Tension in the string \( T \)

Step 1: Understand the Motion

• The 3 kg block hangs vertically and moves downward.
• The 20 kg trolley moves horizontally.
• Both bodies have the same acceleration due to inextensible string.

NEET Point: Connected bodies with a light inextensible string have the same acceleration.

Step 2: Forces on 3 kg Block

Weight:

\[ W = mg = 3 \times 10 = 30\,N \]

Applying Newton's Second Law:

\[ 30 - T = 3a \quad \cdots (1) \]

Step 3: Forces on 20 kg Trolley

Newton's Second Law:

\[ T - f_k = 20a \quad \cdots (2) \]

Step 4: Friction Calculation

\[ N = mg = 20 \times 10 = 200\,N \]

\[ f_k = \mu_k N = 0.04 \times 200 = 8\,N \]

Step 5: Substitute Friction

\[ T - 8 = 20a \quad \cdots (3) \]

Step 6: Solve Equations

From (1):

\[ T = 30 - 3a \]

Substitute into (3):

\[ (30 - 3a) - 8 = 20a \]

\[ 22 = 23a \]

\[ a = \frac{22}{23} = 0.96\,m\,s^{-2} \]

Acceleration: \( a = 0.96\,m\,s^{-2} \)

Step 7: Tension

\[ T = 30 - 3a = 30 - 2.88 = 27.12\,N \]

Tension: \( T = 27.1\,N \)

Final Answer

Acceleration: \( 0.96\,m\,s^{-2} \)

Tension: \( 27.1\,N \)

NEET Quick Revision Points

• Smooth pulley ⇒ same tension throughout string
• Inextensible string ⇒ same acceleration
• \( f_k = \mu_k N \)
• \( N = mg \) (horizontal surface)
• \( F = ma \)
• Always draw FBD and form equations separately