Question

For the following question, enter the correct numerical value up to TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numerical value 5 will be written as 5.00 and 2.346 will be written as 2.35) An elevator and its load have a total mass of 800 kg. If the elevator, originally moving downward at \(10\,\text{m s}^{-1}\), is brought to rest with constant deceleration in a distance of 25 m, the tension in the supporting cable will be _____ N. (Take \( g = 10\,\text{m s}^{-2} \))

Hint:

When an elevator moving downward slows down, its acceleration is upward and cable tension becomes greater than its weight.

Answer 1

Correct Answer: 9600

Step 1: Given data: \[ m = 800\,\text{kg}, \quad u = 10\,\text{m s}^{-1} \ (\text{downward}), \quad v = 0, \] \[ s = 25\,\text{m}, \quad g = 10\,\text{m s}^{-2}. \]
Step 2: Using the equation of motion, \[ v^2 = u^2 + 2as \] \[ 0 = (10)^2 + 2a(25) \Rightarrow a = -2\,\text{m s}^{-2}. \] Negative sign indicates acceleration is upward with magnitude \(2\,\text{m s}^{-2}\).
Step 3: Applying Newton’s second law (upward positive): \[ T - mg = ma \] \[ T = m(g + a) \]
Step 4: Substituting values, \[ T = 800(10 + 2) = 9600\,\text{N}. \] Hence, the tension in the cable is \[ \boxed{9600.00\,\text{N}}. \]