Question

When a person stands on a weighing balance, working on the principle of Hooke's law, it shows a reading of $60kg$ after a long time and the spring gets compressed by $2.5cm$. If the person jumps on the balance from a height of $10cm$, the maximum reading of the balance will be

• A. $60kg$
• B. $120kg$
• C. $180kg$
• D. $240kg$

Answer 1

Answer: (D)

$240kg$

Let the spring constant be k.

$k{x}_0=mg$

$k\times 2.5\times {10}^{-2}=60\times 10$

$k=2400N/m$

$\frac{1}{2}k{x}^2=mg(0.10+x)$

$12000x^2=600(0.10+x)$

$20x^2-x-0.1=0$

$x=\frac{1\pm \sqrt{1+80}}{2\times 20}=0.1$

Now weigh reading is $=kx$

$Mg=kx$

$M\times 10=2400\times 0.1$

$M=240kg$