Question

an elevator and its load weight a total of 800 kg. find the tension T in the supporting cable when the elevator, originally moving downward at 20 ms is brought to rest with constant retardation in a distance of 50m.

Answer 1

Here,

m = 800 kg

u = 20 ms

v = 0 ms

s = 50 m

g = 9.8 ms-2

u2 = -2as

=> 202 = -2a×50

=> a = - 4 ms-2

Now the acceleration acting upwards by the cable = 4 ms-2

So, the elevator experiences an downward acceleration = 4

Net acceleration downwards =  9.8 + 4 = 13.8 ms-2

Tension in the cable will be 800×13.8 = 11040 N 

Answer 2

Answer:

11040 N

Explanation:

Here,

m = 800 kg

u = 20 ms

v = 0 ms

s = 50 m

g = 9.8 ms-2

u2 = -2as

=> 202 = -2a×50

=> a = - 4 ms-2

Now the acceleration acting upwards by the cable = 4 ms-2

So, the elevator experiences an downward acceleration = 4

Net acceleration downwards =  9.8 + 4 = 13.8 ms-2

Tension in the cable will be 800×13.8 = 11040 N