Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

first increases then decreases to zero

decreases

increases

remains same

Step-by-Step Solution

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

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