Given data; Weight of elevator W_{E} = 2500N

Initial velocity u = 0

Distance traveled s = 35m

Time t = 10sec

(1) Since elevator is moving down

Net acceleration force in the down ward direction

= W_{E }– T = (2500 – T)N ...(i)

The net accelerating force produces acceleration ‘a’ in the down ward direction.

Using the relation, F = ma 2500 – T = (2500/9.81)a

T = 2500 – (2500/9.81)a

Hence the above equation represents the general equation for the elevator cable tension when the elevator is moving downward.

(2) Using relation

s = ut + 1/2 at^{2} = 35 = 0 X 10 + 1/2 X a (10)^{2} ...(ii)

∴ a = 0.7 m/sec^{2}

Substituting this value of a in the equation of cable tension

T = 2500 – (2500/9.81) X 0.7T = 2321.61N

(3) T = 2500 – (2500/9.81)a

Limit of cable tension is depends upon the value of a, which varies from 0 to g i.e. 9.81m/sec^{2} At a = 0, T = 2500

i.e elevator freely down

At a = 9.81, T = 0

i.e elevator is at the top and stationary.

**Hence Limits are 0 to 2500N**