Physics, asked by onkareswarisahu, 5 hours ago

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​

Answers

Answered by ss0228896
0

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.

Answered by jeemitvpandya
0

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.

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