A lift, starting from rest, descends with a uniform acceleration of 3 ms–2 until it reaches a speed of 9 ms–1. It then travels at a constant speed of 9 ms–1 for a short time and finally, it is brought to rest with a uniform retardation of 2 ms–2. An object, of mass 6 kg, is on the floor of the lift. Calculate the magnitude of the reaction of the floor on the object during each of the three stages of the motion.
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Here we have to use the Newton’s laws of motion.
During the calculations I will use first two laws of motions.
The first law states that until there is any external force acts on a body it will be at rest or traveling at a constant speed.
In the second law it says that the force applied on a body is proportional to its acceleration. (F = ma)
Let the reaction from the floor is R.
In the first stage the lift goes down with acceleration. So we have to use Newton’s second law.
`darrF = ma`
`R-6g = 6xx3`
`R = 6(g+3) = 76.86N`
In the second stage the lift travels at a constant speed where we have to use Newton’s first law of motion. Since the lift travels constant speed there will be no external forces. The reaction from the floor is same as the weight of the object.
`R = 6g = 58.86N`
In the third stage the lift travels with a deceleration where again we should use Newton’s second law of motion.
`darrF = ma`
`R-6g = 6xx(-2)`
`R = 6(g-2) = 46.86N`
So the reactions on the object from the floor are as follows.
At acceleration stage reaction is 76.86N
At constant speed section reaction is 58.86N
At deceleration section reaction is 46.86N
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