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NEET PHYSICSLAWS OF MOTIONMedium

Question

A man is standing on a spring platform. Reading of spring balance is 60 kg-wt. If the man jumps off the platform, then the reading of the spring balance:

A

first increases then decreases to zero

B

decreases

C

increases

D

remains same

Step-by-Step Solution

  1. Initial State: The man is stationary. The normal reaction (NN) from the spring platform balances his weight (mgmg). Reading = mg=60 kg-wtmg = 60 \text{ kg-wt}.
  2. Jumping Phase (Acceleration): To jump, the man must accelerate upwards. According to Newton's Second Law (Fnet=maF_{net} = ma), the net force must be upwards. Thus, the upward normal force (NN) exerted by the platform must be greater than his weight (mgmg): Nmg=ma    N=m(g+a)N - mg = ma \implies N = m(g + a) [Source 75].
  3. Action-Reaction: By Newton's Third Law, if the platform pushes the man up with force N>mgN > mg, the man pushes the platform down with an equal force N>mgN > mg [Source 57]. This increased downward force compresses the spring further, causing the reading to increase momentarily.
  4. Flight Phase: Once the man's feet leave the platform, the contact force becomes zero (N=0N=0). Consequently, the reading on the spring balance decreases to zero.

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.

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