A 60 kg boy stands on a scale in the elevator. The elevator starts moving and records 450 N. Find the acceleration of the elevator.

Options

A. `2.5 ms^(-2)` upward
B. `2.5 ms^(-2)` downwards
C. `2.5ms^(-2)` in either direction
D. `2.0 ms^(-2) `upward

Answer

To solve the problem, we need to analyze the forces acting on the boy standing on the scale in the elevator.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the boy (W) acting downwards is given by the formula:
\[ W = m \cdot g \]
where \( m = 60 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
Thus,
\[ W = 60 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 588 \, \text{N} \]
- The normal force (N) acting upwards from the scale is given as 450 N.

2. **Apply Newton's Second Law**:
- According to Newton's second law, the net force acting on the boy can be expressed as:
\[ F_{\text{net}} = m \cdot a \]
- The net force can also be expressed in terms of the weight and the normal force:
\[ F_{\text{net}} = W - N \]
- Therefore, we can set up the equation:
\[ W - N = m \cdot a \]

3. **Substitute Known Values**:
- Substituting the values we have:
\[ 588 \, \text{N} - 450 \, \text{N} = 60 \, \text{kg} \cdot a \]
- Simplifying this gives:
\[ 138 \, \text{N} = 60 \, \text{kg} \cdot a \]

4. **Solve for Acceleration (a)**:
- Rearranging the equation to solve for \( a \):
\[ a = \frac{138 \, \text{N}}{60 \, \text{kg}} = 2.3 \, \text{m/s}^2 \]

5. **Determine the Direction of Acceleration**:
- Since the normal force (N) is less than the weight (W), the acceleration is in the downward direction.

### Final Answer:
The acceleration of the elevator is \( 2.3 \, \text{m/s}^2 \) downward.

---