A 24 kg block resting on a floor has a rope tied to its top. The maximum tension , the rope can withstand without breaking is 310 N . The minimum time in which the block can be lifted a vertical distance of 4.6 m by pulling on the rope is

Options

A. `1.2s`
B. `1.3s`
C. `1.7s`
D. `2.3s`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the forces acting on the block
The block has two main forces acting on it:
- The gravitational force (weight) acting downwards, which is given by \( W = mg \).
- The tension in the rope acting upwards, which is given as \( T \).

### Step 2: Calculate the weight of the block
Given:
- Mass of the block \( m = 24 \, \text{kg} \)
- Acceleration due to gravity \( g \approx 10 \, \text{m/s}^2 \)

The weight of the block can be calculated as:
\[ W = mg = 24 \, \text{kg} \times 10 \, \text{m/s}^2 = 240 \, \text{N} \]

### Step 3: Set up the equation of motion
According to Newton's second law, the net force acting on the block is equal to the mass of the block multiplied by its acceleration \( a \):
\[ T - W = ma \]
Substituting the values we have:
\[ T - 240 = 24a \]

### Step 4: Substitute the maximum tension into the equation
The maximum tension \( T \) that the rope can withstand is given as \( 310 \, \text{N} \). Substituting this into the equation:
\[ 310 - 240 = 24a \]
\[ 70 = 24a \]

### Step 5: Solve for acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{70}{24} = \frac{35}{12} \, \text{m/s}^2 \approx 2.92 \, \text{m/s}^2 \]

### Step 6: Use the kinematic equation to find the time
We need to find the minimum time \( t \) to lift the block a vertical distance \( s = 4.6 \, \text{m} \). Using the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ s = \frac{1}{2} a t^2 \]
Rearranging for \( t \):
\[ t^2 = \frac{2s}{a} \]
\[ t = \sqrt{\frac{2s}{a}} \]

### Step 7: Substitute the values to find \( t \)
Substituting \( s = 4.6 \, \text{m} \) and \( a = \frac{35}{12} \, \text{m/s}^2 \):
\[ t = \sqrt{\frac{2 \times 4.6}{\frac{35}{12}}} \]
\[ t = \sqrt{\frac{9.2 \times 12}{35}} = \sqrt{\frac{110.4}{35}} \approx \sqrt{3.15} \approx 1.77 \, \text{s} \]

### Final Answer
The minimum time in which the block can be lifted a vertical distance of 4.6 m is approximately **1.77 seconds**.

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