Work Energy And Power Question 152
When a person stands on a weighing balance, working on the principle of Hooke’s law, it shows a reading of 60 kg after a long time and the spring gets compressed by 2.5 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, $ 60g=kx=k(2.5) $ (i) Let x’ be the maximum compression when the person jumps on the balance, then $ \frac{1}{2}k{x}’^2=60g({x^{‘2}}+10) $
$ \Rightarrow \frac{1}{2}[ \frac{60g}{2.5} ]x{{’}^{2}}=60g(x’+10) $
$ \Rightarrow x{{’}^{2}}=5x’+50 $
$ \Rightarrow x’^{2}-5x’^{2}-50=0 $
Solving for x’, we get $ x’=10cm $ if m kg is the reading, then $ Mg=k(0.1) $ (ii)
From Eqs. (i) and (ii), we get $ m=240kg $
BETA
