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NEET PHYSICSEasy

In an elevator moving vertically up with an acceleration gg, the force exerted on the floor by a passenger of mass MM is:

A

MgMg

B

12Mg\frac{1}{2}Mg

C

Zero

D

2Mg2 Mg

Step-by-Step Solution

  1. Identify Forces: The passenger in the elevator experiences two forces:
  • Gravitational force (Weight) acting downwards: W=MgW = Mg.
  • Normal reaction force exerted by the floor acting upwards: NN.
  1. Apply Newton's Second Law: For a body accelerating upwards with acceleration aa, the net force is upwards. Fnet=N−Mg=MaF_{net} = N - Mg = Ma
  2. Solve for Normal Force (Apparent Weight): N=M(g+a)N = M(g + a)
  3. Substitute Given Acceleration: The problem states the upward acceleration is equal to the acceleration due to gravity (a=ga = g). N=M(g+g)=M(2g)=2MgN = M(g + g) = M(2g) = 2Mg
  4. Conclusion: According to Newton's Third Law, the force exerted by the passenger on the floor is equal in magnitude to the normal reaction NN. Thus, the force is 2Mg2Mg. (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, dealing with apparent weight in accelerated frames).
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