NEET Physics Example 4.9: Block and Trolley System (Step-by-Step)
- Dr.Sanjaykumar Pawar
INTERNAL LINKS
/neet-physics-newtons-laws
/friction-notes-class-11
/tension-in-string-problems
/pulley-system-problems-neet
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/class-11-physics-motion-in-a-line
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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:
- Acceleration of the system \( a \)
- 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.
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):
\[ (30 - 3a) - 8 = 20a \]
\[ 22 = 23a \]
\[ a = \frac{22}{23} = 0.96\,m\,s^{-2} \]
Step 7: Tension
\[ T = 30 - 3a = 30 - 2.88 = 27.12\,N \]
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
