Question

Suppose 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.

Answer 1

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
\[\Rightarrow x = \frac{23 . 6}{100} = 0 . 236 \approx 0 . 24 \text{ m }\]
When 1 kg body is added,
total mass = (2 + 1) kg = 3 kg
Let elongation be x'.
∴ kx' = 3g + 3a = 3 × 9.8 + 6

\[\Rightarrow x' = \frac{35 . 4}{100}\]
\[ = 0 . 354 \approx 0 . 36 m\]

So, further elongation = x' − x
                                  = 0.36 − 0.24 = 0.12 m.