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 step by step, we will use the concepts of weight, apparent weight, and Newton's second law of motion.

### Step 1: Identify the given information
- Mass of the boy (m) = 60 kg
- Apparent weight (R') = 450 N
- Acceleration due to gravity (g) = 10 m/s² (approximately)

### Step 2: Calculate the actual weight of the boy
The actual weight (W) of the boy can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 60 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 600 \, \text{N} \]

### Step 3: Set up the equation for apparent weight
When the elevator is accelerating downwards, the apparent weight (R') is given by:
\[ R' = W - m \cdot a \]
Where:
- \( a \) is the acceleration of the elevator.

### Step 4: Substitute the known values into the equation
We know:
- \( R' = 450 \, \text{N} \)
- \( W = 600 \, \text{N} \)

Substituting these into the equation:
\[ 450 = 600 - 60 \cdot a \]

### Step 5: Rearrange the equation to solve for acceleration (a)
Rearranging the equation gives:
\[ 60 \cdot a = 600 - 450 \]
\[ 60 \cdot a = 150 \]

### Step 6: Solve for acceleration (a)
Dividing both sides by 60:
\[ a = \frac{150}{60} \]
\[ a = 2.5 \, \text{m/s}^2 \]

### Step 7: Determine the direction of acceleration
Since the apparent weight is less than the actual weight, the elevator is accelerating downwards. Therefore, the acceleration of the elevator is:
\[ a = 2.5 \, \text{m/s}^2 \, \text{downwards} \]

### Final Answer
The acceleration of the elevator is \( 2.5 \, \text{m/s}^2 \) downwards.

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