To solve the problem, we will use the equations of motion and the concept of forces acting on the elevator. Here’s a step-by-step solution:
### Step 1: Calculate the weight of the elevator
The weight (W) of the elevator can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 1600 \, \text{kg} \) (mass of the elevator)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
Calculating the weight:
\[ W = 1600 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 15680 \, \text{N} \]
### Step 2: Determine the net force acting on the elevator
The net force (F_net) acting on the elevator as it falls can be found by subtracting the retarding force from the weight:
\[ F_{\text{net}} = W - F_{\text{retarding}} \]
where:
- \( F_{\text{retarding}} = 3700 \, \text{N} \)
Calculating the net force:
\[ F_{\text{net}} = 15680 \, \text{N} - 3700 \, \text{N} = 11980 \, \text{N} \]
### Step 3: Calculate the acceleration of the elevator
Using Newton's second law, we can find the acceleration (a) of the elevator:
\[ F_{\text{net}} = m \cdot a \]
Rearranging gives:
\[ a = \frac{F_{\text{net}}}{m} \]
Calculating the acceleration:
\[ a = \frac{11980 \, \text{N}}{1600 \, \text{kg}} \approx 7.49 \, \text{m/s}^2 \]
### Step 4: Use the equations of motion to find the final speed
We will use the third equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity (what we want to find)
- \( u = 0 \, \text{m/s} \) (initial velocity, since it starts from rest)
- \( a = 7.49 \, \text{m/s}^2 \) (acceleration we calculated)
- \( s = 72 \, \text{m} \) (displacement)
Substituting the values:
\[ v^2 = 0 + 2 \cdot 7.49 \, \text{m/s}^2 \cdot 72 \, \text{m} \]
\[ v^2 = 1078.08 \]
\[ v = \sqrt{1078.08} \approx 32.8 \, \text{m/s} \]
### Final Answer
The speed at which the elevator hits the bottom of the shaft is approximately **32.8 m/s**.
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