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NEET PHYSICSLAWS OF MOTIONMedium

Question

Calculate the acceleration of the block and trolley system shown in the figure. The coefficient of kinetic friction between the trolley and the surface is 0.050.05. (Take g=10 m/s2g=10 \text{ m/s}^2, the mass of the string is negligible and no other friction exists). [Note: Based on the answer, the mass of the hanging block is 2 kg2 \text{ kg} and the mass of the trolley is 10 kg10 \text{ kg}].

A

1.25 m/s²

B

1.50 m/s²

C

1.66 m/s²

D

1.00 m/s²

Step-by-Step Solution

  1. Analyze the System: The system consists of a trolley of mass MM (typically 10 kg10 \text{ kg} in this specific PYQ) on a horizontal surface and a block of mass mm (typically 2 kg2 \text{ kg}) hanging vertically, connected by a string over a pulley.
  2. Identify Forces:
  • For the hanging block (mm): Gravity acts downwards (mgmg) and tension (TT) acts upwards. Equation of motion: mgT=mamg - T = ma.
  • For the trolley (MM): Tension (TT) acts forward and kinetic friction (fkf_k) acts backward. Equation of motion: Tfk=MaT - f_k = Ma.
  1. Calculate Friction: The normal reaction on the trolley is N=MgN = Mg. The kinetic friction is fk=μkN=μkMgf_k = \mu_k N = \mu_k Mg [Source 127, 134].
  2. Solve for Acceleration (aa): Adding the two equations eliminates tension TT: mgfk=(M+m)amg - f_k = (M + m)a mgμkMg=(M+m)amg - \mu_k Mg = (M + m)a a=g(mμkM)M+ma = \frac{g(m - \mu_k M)}{M + m} [Source 129]
  3. Calculation: Substituting the standard values for this problem (m=2 kg,M=10 kg,μk=0.05,g=10 m/s2m=2 \text{ kg}, M=10 \text{ kg}, \mu_k=0.05, g=10 \text{ m/s}^2): a=10(20.05×10)10+2a = \frac{10(2 - 0.05 \times 10)}{10 + 2} a=10(20.5)12=10(1.5)12=1512=1.25 m/s2a = \frac{10(2 - 0.5)}{12} = \frac{10(1.5)}{12} = \frac{15}{12} = 1.25 \text{ m/s}^2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONcalculateaccelerationtrolleysystemfigure

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