Question

A machine lifts a load of 500 Kgf is a vertically upward direction. it this machine is driven by a forcely falling mass of 100 kg through a distance of 4m is 2s, find: (Take g = 10 m/s²) (a) force exerted by the falling mass​

Answer 1

Answer:

=0

As the lift is freely falling, thus its acceleration in down ward direction a=g

Let the mass of the body be m and force acting on it due to spring be F.

Writing equation of motion for the body : ma=mg−F

Or mg=mg−F

⟹ F=0

Thus no force is acting on the body due to spring, so the reading of the spring balance is zero.

Answer 2

Answer:

Correct option is

A

=0

As the lift is freely falling, thus its acceleration in down ward direction a=g

Let the mass of the body be m and force acting on it due to spring be F.

Writing equation of motion for the body : ma=mg−F

Or mg=mg−F

⟹ F=0

Thus no force is acting on the body due to spring, so the reading of the spring balance is zero.

solution