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An elevator and its load weigh a total o...

An elevator and its load weigh a total of 1600 kg. Find the tension T in the supporting cable when the elevator, originally moving downward at 20 m/s is brought to rest with constant acceleration in a distance of 50 m.

A

`2024xx10^(2)N`

B

`1024xx10^(4)N`

C

`2024xx10^(4)N`

D

`2024xx10^(-4)N`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the tension \( T \) in the supporting cable of an elevator that is brought to rest, we can follow these steps: ### Step 1: Identify the Given Data - Total weight of the elevator and load, \( W = 1600 \, \text{kg} \) - Initial velocity, \( u = 20 \, \text{m/s} \) (downward) - Final velocity, \( v = 0 \, \text{m/s} \) (at rest) - Displacement, \( s = 50 \, \text{m} \) ### Step 2: Use the Third Equation of Motion We can use the third equation of motion to find the acceleration \( a \): \[ v^2 = u^2 + 2as \] Substituting the known values: \[ 0 = (20)^2 + 2a(50) \] \[ 0 = 400 + 100a \] Rearranging gives: \[ 100a = -400 \] \[ a = -4 \, \text{m/s}^2 \] ### Step 3: Identify the Direction of Acceleration Since the elevator is moving downward and is being brought to rest, the acceleration \( a \) is upward. Thus, the effective acceleration acting on the elevator is: \[ a = 4 \, \text{m/s}^2 \, \text{(upward)} \] ### Step 4: Calculate the Weight of the Elevator The weight \( W \) of the elevator is given by: \[ W = mg = 1600 \times 9.8 = 15680 \, \text{N} \] ### Step 5: Apply Newton's Second Law According to Newton's second law, the net force acting on the elevator can be expressed as: \[ T - W = ma \] Where: - \( T \) is the tension in the cable, - \( W \) is the weight of the elevator, - \( m \) is the mass of the elevator, - \( a \) is the acceleration. Substituting the values we have: \[ T - 15680 = 1600 \times 4 \] \[ T - 15680 = 6400 \] ### Step 6: Solve for Tension \( T \) Rearranging the equation to find \( T \): \[ T = 15680 + 6400 \] \[ T = 22080 \, \text{N} \] ### Final Answer The tension in the supporting cable when the elevator is brought to rest is: \[ T = 22080 \, \text{N} \] ---
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