Figure -8413. The elevator shown in figure (5-E5) is descending witban

1.	Figure -8413. The elevator shown in figure (5-E5) is descending witban acceleration of 2 m/'s. The mass of the block A is05 kg. What force is exerted by the block A os theblock B?2 m/s261Figure 5-E5
Answer» Answer:If the elevator was not moving the force applied by the block A on the block B would have been equal to its weight = mg = 0.5 kg x 9.8 m/s²= 4.9 N. But in this case the elevator is descending with an acceleration of a=2 m/s². So in order to apply the Newton's Laws of motion with respect to the elevator we apply a pseudo force equal to ma on the block A in the direction opposite to the acceleration ie upwards. The net force on the block is = mg-ma= 4.9 - 0.5x2 = 4.9-1 = 3.9 N ≈ 4.0 N

Figure -8413. The elevator shown in figure (5-E5) is descending witban acceleration of 2 m/'s. The mass of the block A is05 kg. What force is exerted by the block A os theblock B?2 m/s261Figure 5-E5

Answer»

Answer:If the elevator was not moving the force applied by the block A on the block B would have been equal to its weight = mg = 0.5 kg x 9.8 m/s²= 4.9 N. But in this case the elevator is descending with an acceleration of a=2 m/s². So in order to apply the Newton's Laws of motion with respect to the elevator we apply a pseudo force equal to ma on the block A in the direction opposite to the acceleration ie upwards. The net force on the block is

= mg-ma= 4.9 - 0.5x2 = 4.9-1 = 3.9 N ≈ 4.0 N

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