Question

An elevator and its load have a total mass of 800 kg. Find the tension T in the supporting cable when the elevator, originally moving downward at 10m/s is brought to the rest with constant acceleration at a distance of 25m.

Answer 1

Mass Of The Elevator=800 kg

Tension T Supporting The cable
Intial Speed of Elevator (u)=10 m/s
Final Speed of Elevator (v)=0 m/s
Distance (s)=25 m

WE have to Find Accerlation
Use Equation of Motion
v²=u²+2as
0²=10²+2×a×25
0-10²=50a
-100=50a
a=-100/50
a=-2 m/s²

MEANS Accerlation (Retardation)Needed 2m/s² to Take Elevator at rest.
Now We know
Accerlation=2m/s². to Come rest.

Now On Cable While Coming It have Also Take Gravitational Force of=10 m/s
MEANS Total Accerlation=10+2=12 m/s²

Now Tension in cable
=Mass×Total Accerlation
=800×12
=9600N

Answer 2

Total mass of elevator and its load = 800 kg

Tension in the cable = T

Gravitiaonal force= mg

Refer to the FBD,

Apply Newton's law,

T - mg = ma

or, T = m(g + a) ...(i)

Now we are required to find the acceleration,

here, Intital velocity = u = - 10 m/s

and, final velocity = v = 0

distance = S = -25 m

v² + u² = 2aS

or, a =  - u² / 2y = (-10)² /2*(- 25) = 2 m/s²

Substituting value of acceleration in eq(i):

T = m(g+a)

or, T = 800 (9.8 + 2)

or, T = 800 * 11.8

or, T = 9440 N

Therefore, Tension in the supporting cable is 9440 N