Question

2.
A lift of mass m is supported by a cable that can
with stand a force of 3mg. Find the shortest
distance in which the lift can be stopped when it is
descending with a speed of g/4.
1) g/8 m 2)g/16 m 3) g/32 m 4) g/64 m​

Answer 1

answer : option (4) g/64

it is given that a lift of mass m is supported by a cable that can with stand a force of 3mg.

we have to find the shortest distance in which the lift can be stopped when it is descending with a speed of g/4.

so, T = mg + ma

⇒3mg = mg + ma

⇒a = 2g , it means acceleration is directed upward.

now using formula, v² = u² + 2as

let shortest distance is s in which lift can be stopped.

here final velocity, v = 0, u = g/4 , a = 2g

so, 0 = (g/4)² + 2(-2g)s

⇒g²/16 = 4gs

⇒s = g/64

therefore, shortest distance in which lift can be stopped when it is descending with speed of g/4 is g/64.

also read similar questions : How much momentum will an object of mass 10 kg transfer to the floor if it falls from a height of 5 m? (g=10m/s)

https://brainly.in/question/15630706

A person of mass 60 kg is inside a lift of mass 940 kg and posses the button on control panel . The lift starts moving u...

https://brainly.in/question/1340197