A 71 kg man stands on a bathroom scale in an elevator. What does the scale read if the elevator is ascending with an acceleration of `3.0" m"//"s"^(2)` ?

Options

A. 140 N
B. 480 N
C. 690 N
D. 910 N

Answer

To find out what the scale reads when the elevator is ascending with an acceleration of \(3.0 \, \text{m/s}^2\), we can follow these steps:

### Step 1: Understand the Forces Acting on the Man
When the man is standing on the scale in the elevator, two forces act on him:
1. The gravitational force (weight) acting downwards, given by \(W = mg\), where \(m\) is the mass of the man and \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).
2. The normal force (\(N\)) exerted by the scale acting upwards.

### Step 2: Calculate the Gravitational Force
Using the mass of the man (\(m = 71 \, \text{kg}\)):
\[ W = mg = 71 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 695.8 \, \text{N} \]

### Step 3: Apply Newton's Second Law
Since the elevator is accelerating upwards, we can apply Newton's second law:
\[ N - W = ma \]
Where:
- \(N\) is the normal force (what the scale reads),
- \(W\) is the weight of the man,
- \(m\) is the mass of the man,
- \(a\) is the acceleration of the elevator.

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

### Step 5: Substitute the Values
Substituting the known values into the equation:
- \(W = 695.8 \, \text{N}\)
- \(m = 71 \, \text{kg}\)
- \(a = 3.0 \, \text{m/s}^2\)

Calculating \(ma\):
\[ ma = 71 \, \text{kg} \times 3.0 \, \text{m/s}^2 = 213 \, \text{N} \]

Now substituting back into the equation for \(N\):
\[ N = 695.8 \, \text{N} + 213 \, \text{N} = 908.8 \, \text{N} \]

### Step 6: Conclusion
The scale reads \(908.8 \, \text{N}\).