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Two masses, 2 kg and 3 kg are placed on the floor of a lift as shown in the figure below. The force of normal reaction applied by a 3 kg block on the floor of the lift is 40 N. Which of the following statements is/are correct? [$g = 10m/{s^2}$ ]
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A. The lift is moving with an upward acceleration.
B. The lift is moving with a downward acceleration
C. The lift is either at rest of moving with uniform speed
D. Force of normal reaction applied by the 2 kg block on the 3 kg block is 16 N.

Last updated date: 24th Jun 2024
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Hint: while we travel in a lift that goes down we feel that we have lost some weight and when lift goes up we feel weight gain. But the actual mass of us is the same in both cases. It's all due to the variation of normal reactions in both the cases. This can be solved from the ground reference frame by using pseudo force which causes variation in normal force.
Formula used:
  & mg - N = ma \cr
  & N = mg - ma \cr
  & N = W - ma \cr} $

Complete answer:
When lift is not moving let the weight of the blocks be W
$W = mg$
Where m is the mass of two blocks combined and g is the acceleration due to gravity
Now when lift starts moving down with acceleration ‘a’
Normal reaction N acts upward
Weight W acts downward
Force $m \times a$ acts downward
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From ground frame of reference
Balancing the forces gives us
  & mg - N = ma \cr
  & \Rightarrow N = mg - ma \cr
  & \Rightarrow N = W - ma \cr
  & \Rightarrow 40 = 5g - 5a \cr
  & \Rightarrow 40 = 50 - 5a \cr
  & \therefore a = 2m/{s^2} \cr} $
So the lift is moving down with acceleration 2 meter per second square
Now normal reaction between 2kg and 3kg block will be
  & {m_2}g - N = {m_2}a \cr
  & \Rightarrow N = W - {m_2}a \cr
  & \Rightarrow N = 2g - 2a \cr
  & \Rightarrow N = 20 - 4 \cr
  & \therefore N = 16newton \cr} $

Hence options B and D are correct.

Additional information:
Same question can be solved by the Lift frame of reference. Here pseudo force comes into action. Since lift blocks system is moving down with acceleration ‘a’ pseudo force will be acting upward and it will be $m \times a$
Upward forces = $N + ma$
Downward forces = $mg$
Equating both the forces we get
$N = W - ma$
Hence both methods will give us the same answer

It is to be noted that only weight seems to be increased but not mass. Because mass is always constant wherever we go but weights vary as resultant acceleration varies. A freely falling body feels no weight due to the same property as resultant acceleration would be (g-g =0) zero. This is called weightlessness. Same applies with the astronaut in the satellite where he feels weightlessness.