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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.`
one year ago

Arun
23546 Points
```							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/s²F = 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 velocitya = acceleration of the lift = 0 m/s²W - F = 0 F = W F = 147 N
```
one year ago
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