Asked in ETEA MDCAT 2012 2012Moderate

Two blocks of masses 1.0 kg and 3.0 kg placed in contact are acted upon by a force of 40 N. The acceleration of 1.0 kg mass will be:

Correct answer: B. 10 ms^-2

  • A. 40 ms^-2
  • B. 10 ms^-2
  • C. 30 ms^-2
  • D. 50 ms^-2

Explanation

In this scenario, the two blocks are in contact and can be treated as a single system with a total mass of 1.0 kg + 3.0 kg = 4.0 kg. The force applied is 40 N. According to Newton's second law, F = ma, where F is the force applied, m is the total mass, and a is the acceleration.Here, the acceleration a can be calculated as:a = F / (m1 + m2) = 40 N / 4.0 kg = 10 ms^-2.Thus, the acceleration of the 1.0 kg mass is 10 ms^-2, as both blocks will accelerate together as a single system.Option A (40 ms^-2) is incorrect as it does not account for the total mass. Option C (30 ms^-2) is a miscalculation. Option D (50 ms^-2) exceeds the possible acceleration given the force and mass.

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