Question

A mass $m$ hung on a vertical ideal spring (of spring constant $k$) is initially held at its highest position above equilibrium such that the spring is compressed by $x_1$. When released, the mass falls through a distance $2h$ such that the lowest point it reaches is when the spring is stretched by $x_2$.
Based on this, what is $x_2-x_1$? Neglect energy loss through friction or air resistance.

• A. $0$
• B. $h$
• C. $2h$
• D. $\frac{mg}{k}$
• E. $\frac{2mg}{k}$

Answer 1

The correct option is E $\frac{2mg}{k}$

As at the highest and lowest position of the spring the velocity of the mass is zero, thus initial and final kinetic energy of the mass is zero i.e $\Delta K.E=0$

From figure $x_1+x_2=2h$ .........(1)

Work done by spring force $W_S=-\Delta P.E=-\frac{1}{2}k(x_2^2-x_1^2)$

Work done by gravity $W_g=mg\left(2h\right)$

Using work-energy theorem : $W_{net}=\Delta K.E$

$\therefore$ $-\frac{1}{2}k(x_2^2-x_1^2)+2mgh=0$

OR $(x_2^2-x_1^2)k=4mgh$

OR $(x_2-x_1)(x_2+x_1)k=4mgh$

OR $(x_2-x_1)\left(2h\right)k=4mgh$ $⟹x_2-x_1=\frac{2mg}{k}$