A lift of mass 100 kg is moving upwards with an acceleration of 1 `m//s^(2)`. The tension developed in the string, which is connected to lift is (`g=9.8m//s^(2)`)

Options

A. 980 N
B. 1080 N
C. 1100 N
D. 1000 N

Answer

To find the tension in the string connected to a lift moving upwards with an acceleration, we can follow these steps:

### Step 1: Identify the Forces Acting on the Lift
- The lift has a mass \( m = 100 \, \text{kg} \).
- The gravitational force acting downwards on the lift is given by \( F_g = m \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \).
- The tension \( T \) in the string acts upwards.

### Step 2: Calculate the Gravitational Force
\[ F_g = m \cdot g = 100 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 980 \, \text{N} \]

### Step 3: Apply Newton's Second Law
According to Newton's second law, the net force acting on the lift can be expressed as:
\[ \text{Net Force} = T - F_g = m \cdot a \]
Where:
- \( a = 1 \, \text{m/s}^2 \) (the upward acceleration of the lift).

### Step 4: Set Up the Equation
Substituting the known values into the equation:
\[ T - 980 \, \text{N} = 100 \, \text{kg} \cdot 1 \, \text{m/s}^2 \]
\[ T - 980 \, \text{N} = 100 \, \text{N} \]

### Step 5: Solve for Tension \( T \)
Rearranging the equation to solve for \( T \):
\[ T = 980 \, \text{N} + 100 \, \text{N} = 1080 \, \text{N} \]

### Final Answer
The tension developed in the string is \( T = 1080 \, \text{N} \).

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