Question

A mass of $2\text{kg}$ is released from a height of $40\text{cm}$ on a spring whose force constant of $1960\text{N/m}$. The maximum compression in the spring is

• A. $10\text{cm}$
• B. $1\text{cm}$
• C. $20\text{cm}$
• D. $5\text{cm}$

Answer 1

The correct option is A $10\text{cm}$

Given:
$m=2\text{kg};h=40\text{cm};k=1960\text{N/m}$

Here, all forces acting on the system are conservative in nature, so the total energy $E$ of the system will be conserved.

$\therefore E=\text{constant}$

$\Rightarrow \Delta E=0$

$\Rightarrow \Delta U+\Delta KE=0$

There are two types of potential energy present in this scenario, one is spring potential energy and the other one is gravitational potential energy.

$\therefore {\left(\Delta U\right)}_{spring}+{\left(\Delta U\right)}_{grav}+\Delta KE=0$

Looking at initial and final situation, kinetic energy is zero in both cases.

$\because \Delta KE=0$

$\therefore {\left(\Delta U\right)}_{spring}+{\left(\Delta U\right)}_{grav}=0$

Let $x$ be the maximum compression on the spring. So

Decrease in gravitational potential energy = increase in elastic potential energy

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

$2\times 9.8(0.4+x)=\frac{1}{2}\times 1960\times x^2$

$1960x^2-39.2x-15.68=0$

After solving this quadratic equation, we get

$x=0.10or-0.08$

Since, compression occurs, so we will take positive value only

$\therefore x=0.1\text{m}=10\text{cm}$

Hence, option $\left(A\right)$ is the correct choice.

Why this question ? If no external forces are acting (or the work done by them is zero) on the system, and the internal forces are conservative, then the mechanical energy of the system remains constant.