CBSE-XI-Physics
05: Newton's Laws of Motion
- #18Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s2. Find the elongation.Ans : When the ceiling of the elevator is going up with an acceleration 'a', then a pseudo-force acts on the block in the downward direction.
a = 2 m/s2
From the free-body diagram of the block,
kx = mg + ma
⇒ kx = 2g + 2a
= 2 × 9.8 + 2 × 2
= 19.6 + 4
`` \Rightarrow x=\frac{23.6}{100}=0.236\approx 0.24\,\mathrm{\,m \,}``
When 1 kg body is added,
total mass = (2 + 1) kg = 3 kg
Let elongation be x'.
∴ kx' = 3g + 3a = 3 × 9.8 + 6
`` \Rightarrow x\text{ ' }=\frac{35.4}{100}``
`` =0.354\approx 0.36\,\mathrm{\,m \,}``
So, further elongation = x' - x
= 0.36 - 0.24 = 0.12 m.
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