An elevator is moving vertically up with a acceleration a the force exerted on the floor by a passenger of mass m is

Options

A. `1/2 mg`
B. mg
C. mg+ma
D. Zero

Answer

To solve the problem of finding the force exerted on the floor by a passenger of mass \( m \) in an elevator moving vertically upwards with acceleration \( a \), we can follow these steps:

### Step 1: Identify the Forces Acting on the Passenger
In the elevator, the passenger experiences two main forces:
1. The gravitational force (weight) acting downward, which is given by \( F_g = mg \), where \( g \) is the acceleration due to gravity.
2. The normal force \( N \) exerted by the floor of the elevator acting upward.

### Step 2: Consider the Frame of Reference
Since the elevator is accelerating upwards with acceleration \( a \), we can analyze the forces from the perspective of the passenger inside the elevator. In this non-inertial frame, we need to account for a pseudo force acting downward due to the upward acceleration of the elevator.

### Step 3: Apply Newton’s Second Law
In the upward accelerating frame of reference, the net force acting on the passenger can be expressed as:
\[ N - mg - ma = 0 \]
Here, \( ma \) is the pseudo force acting downward.

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ N = mg + ma \]

### Step 5: Factor Out the Mass
We can factor out \( m \) from the right-hand side:
\[ N = m(g + a) \]

### Conclusion
Thus, the force exerted on the floor by the passenger of mass \( m \) is:
\[ N = m(g + a) \]

### Final Answer
The force exerted on the floor by a passenger of mass \( m \) is \( m(g + a) \).

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