Question

An elevator cab and its load have a combined mass of 1200 kg. Find the tension in the supporting cable when the cab, originally moving downward at 10 m/s, is brought to rest with constant acceleration in a distance of 35 m.

Answer 1

10044 N

Step-by-step explanation:

The acceleration of the cab is calculated using the equation of motion:

`v^2 = u^2+2as`

v is the final velocity = 0 m/s in this question, since it is brought to rest

u is the initial velocity = 10 m/s

a is the acceleration

s is the distance = 35 m

`a = (v^2-u^2)/(2s) = \frac{(0 \text{ m/s})^2-(10 \text{ m/s})^2}{2* (35\text{ m})} = -1.43\text{ m/s}^2`

Since it accelerates downwards, its resultant acceleration is

`a_R = g + a`

g is the acceleration of gravity.

`a_R = (9.8-1.43)\text{ m/s}^2 = 8.37\text{ m/s}^2`

The tension in the cable is

`T = ma_R = (1200\text{ kg})(8.37\text{ m/s}^2) = 10044 \text{ N}`