SATHEE: Work Energy And Power Question 152

Work Energy And Power Question 152

Question: When a person stands on a weighing balance, workingon the principle of Hooke’s law, it shows a reading of 60 kg after a long time and the spring gets compressed by2.5 cm. If the person jumps on the balance from a height of 10 cm, the maximum reading of the balance will be

Options:

A) 60 kg

B) 120kg

C) 180 kg

D) 240kg

Show Answer

Answer:

Correct Answer: D

Solution:

[d] initially, $60 g = k x = k (2.5)$ (i) Let x’ be the maximum compression when the person jumps on the balance, then $\frac{1}{2} k x^{'} = 60 g (x^{‘ 2} + 10)$

$\Rightarrow \frac{1}{2} [\frac{60 g}{2.5}] x {^{'}}^{2} = 60 g (x^{'} + 10)$

$\Rightarrow x {^{'}}^{2} = 5 x^{'} + 50$

$\Rightarrow x {^{'}}^{2} - 5 x {^{'}}^{2} - 50 = 0$

Solving for x’, we get $x^{'} = 10 c m$ if m kg is the reading, then $M g = k (10)$ (ii)

From Eqs. (i) and (ii), we get $m = 240 k g$

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Work Energy And Power Question 152

Question: When a person stands on a weighing balance, workingon the principle of Hooke’s law, it shows a reading of 60 kg after a long time and the spring gets compressed by2.5 cm. If the person jumps on the balance from a height of 10 cm, the maximum reading of the balance will be

Options:

Answer:

Solution:

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