In an elevator moving vertically up with an acceleration $g$, the force exerted on the floor by a passenger of

Question

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

Accepted Answer

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

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