Question

 An elevator weighing 12 kN starts from rest and acquires an upward velocity of 2 m/sec in a distance of 5 m. If the acceleration is constant, what is the tension in the cable? *

Answer 1

We know that

Newton motion third

Law

$V^2=U^2+2aS$

$a=\frac{V^2-U^2}{2S}$

Put value

$a=\frac{2^2-0^2}{2\times5}=0.4m/sec^2$

Now We know that

$T-ma=mg$

$$T=mg(1+\frac{a}{g})=12000\times(1+\frac{0.4}{9.8})=12.489KN$$