A certain crane can provide a maximum lifting force 2500 N. It hoists a 2000 kg load starting at ground level by applying the maximum force for a 2 second interval, then, it applies just sufficient force to keep the load moving upward at constant speed. Approximately how long does it take to raise the load from ground level to a height of 30 m ?

Options

A. 2 s
B. 7 s
C. 5 s
D. 9 s

Answer

To solve the problem of how long it takes for the crane to raise a 2000 kg load to a height of 30 m, we will break the solution into clear steps.

### Step 1: Calculate the weight of the load
The weight of the load (W) can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 2000 \, \text{kg} \) (mass of the load)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 2000 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20000 \, \text{N} \]

### Step 2: Determine the net force during the initial lifting phase
The crane can provide a maximum lifting force of 2500 N. The net force (F_net) acting on the load when the crane is applying maximum force can be calculated as:
\[ F_{\text{net}} = F_{\text{applied}} - W \]
where:
- \( F_{\text{applied}} = 2500 \, \text{N} \)

Calculating the net force:
\[ F_{\text{net}} = 2500 \, \text{N} - 20000 \, \text{N} = -17500 \, \text{N} \]

Since the applied force is less than the weight, the crane cannot lift the load initially. Hence, we need to consider the crane's maximum lifting force.

### Step 3: Calculate the acceleration of the load
When the crane applies the maximum force for 2 seconds, we need to find the net force and then the acceleration (a):
\[ F_{\text{net}} = F_{\text{applied}} - W \]
Using the maximum lifting force:
\[ F_{\text{net}} = 2500 \, \text{N} - 20000 \, \text{N} = -17500 \, \text{N} \]

Since the crane cannot lift the load, we need to consider the scenario where it can apply enough force to lift the load.

### Step 4: Calculate the acceleration while lifting
If the crane applies a force greater than the weight, we can recalculate:
Assuming the crane can apply a force sufficient to lift the load, we can find the net force and acceleration:
\[ F_{\text{net}} = 2500 \, \text{N} - 20000 \, \text{N} = -17500 \, \text{N} \]

This indicates that the crane cannot lift the load with the given parameters.

### Step 5: Calculate the speed after 2 seconds
Using the first equation of motion:
\[ v = u + at \]
where:
- \( u = 0 \, \text{m/s} \) (initial velocity)
- \( a = \frac{F_{\text{net}}}{m} = \frac{2500 \, \text{N}}{2000 \, \text{kg}} = 1.25 \, \text{m/s}^2 \) (acceleration)
- \( t = 2 \, \text{s} \)

Calculating the final velocity after 2 seconds:
\[ v = 0 + (1.25 \, \text{m/s}^2 \cdot 2 \, \text{s}) = 2.5 \, \text{m/s} \]

### Step 6: Calculate the distance covered in 2 seconds
Using the formula for distance:
\[ s = ut + \frac{1}{2} a t^2 \]
Calculating the distance:
\[ s = 0 + \frac{1}{2} \cdot 1.25 \, \text{m/s}^2 \cdot (2 \, \text{s})^2 = 5 \, \text{m} \]

### Step 7: Calculate the remaining distance and time
The total height to be lifted is 30 m. After 2 seconds, the crane has lifted the load 5 m, so the remaining distance is:
\[ 30 \, \text{m} - 5 \, \text{m} = 25 \, \text{m} \]

Now, the crane will move at a constant speed of 2.5 m/s for the remaining distance. The time taken to cover this distance can be calculated using:
\[ t = \frac{\text{distance}}{\text{speed}} = \frac{25 \, \text{m}}{2.5 \, \text{m/s}} = 10 \, \text{s} \]

### Step 8: Calculate the total time taken
The total time taken to lift the load is:
\[ \text{Total time} = 2 \, \text{s} + 10 \, \text{s} = 12 \, \text{s} \]

### Final Answer
The total time taken to raise the load from ground level to a height of 30 m is approximately **12 seconds**.

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