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A body of mass 15 kg is hung by a spring balance in a lift. what would be the reading of the balance when : (i) ascending with an acceleration 2m/s^2. (ii) descending with a constant velocity of 2m/s.

  1. A body of mass 15 kg is hung by a spring balance in a lift. what would be the reading of the balance when : (i) ascending with an acceleration 2m/s^2. (ii) descending with a constant velocity of 2m/s.

Grade:11

1 Answers

Arun
25750 Points
5 years ago
1)
m = mass of the body hanged from spring balance = 15 kg 
W = actual weight of the body hanged in downward direction = mg = 15 x 9.8 = 147 N 
a = acceleration of the lift in upward direction = 2 m/s²
F = reading of the spring balance 
force equation for the motion of the body hanged in the lift is given as 
F - W = ma 
F - 147 = 15 (2)
F = 177 N 

2)
Since the lift is going at constant velocity
a = acceleration of the lift = 0 m/s²
W - F = 0 
F = W 
F = 147 N 
 

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