Animated Solution for Physics - Work, Energy, and Power: An ideal spring with spring constant $k$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is

When you first look at this problem, it seems almost too easy. A block is dropped on a spring, and we need to find how far it stretches. Your brain might immediately jump to the most familiar equation involving springs and masses: $Mg=kx$. You quickly solve for $x$ and get $x=\frac{Mg}{k}$. You look at the options, and there it is—option (c). You mark it and move on, feeling confident.

But wait! If you did this, you just fell into one of the most classic traps in physics.

The equation $Mg=kx$ describes the equilibrium position. This is the point where the upward pull of the spring exactly balances the downward pull of gravity. If you were to hold the block and lower it very, very slowly until you didn't feel its weight anymore, it would rest at this position.

However, the problem states that the mass is released. It is dropped! As it falls towards the equilibrium position, it loses gravitational potential energy and gains kinetic energy. By the time it reaches the equilibrium point, it is moving quite fast. Because of its inertia, it doesn't just stop there; it overshoots the equilibrium position and keeps stretching the spring until all its kinetic energy is drained.

We are looking for the maximum extension, not the equilibrium position. To find this, we need a more powerful tool: the Principle of Conservation of Mechanical Energy.

From Rest to Rest

Imagine the sequence of events.

State 1 (The Release): The block is attached to the unstretched spring. You are holding it. Its velocity is zero ($v_i=0$). This is our starting point.

State 2 (The Maximum Extension): You let go. The block accelerates downwards, zips past the equilibrium point, and continues to stretch the spring. The spring pulls back harder and harder until, finally, the block comes to a momentary halt. At this exact lowest point, its velocity is zero again ($v_f=0$). Let's call this maximum downward displacement $x$.

Conservation of Energy

In this system, there are only two forces doing work on the block: 1. Gravity: A conservative force. 2. Spring Force: Another conservative force.

Since there are no non-conservative forces (like friction or air resistance) draining energy from the system, the total mechanical energy must remain constant.

Mechanical energy is the sum of Kinetic Energy ($K$) and Potential Energy ($U$). Here, we have two types of potential energy: Gravitational ($U_g$) and Elastic/Spring ($U_s$).

Initial and Final Energies

To make our calculations elegant, we get to choose our reference level for gravitational potential energy. Let's set the initial position of the block as our zero-potential line ($U_g=0$).

Analyzing the Initial State: - The block is at rest: $K_i=0$ - It is at our reference level: $U_{gi}=0$ - The spring is unstretched: $U_{si}=0$ - Total Initial Energy: $E_i=0+0+0=0$

Analyzing the Final State (at maximum extension $x$): - The block has momentarily stopped: $K_f=0$ - It has fallen a distance $x$ below the reference level: $U_{gf}=-Mgx$ - The spring is stretched by a distance $x$: $U_{sf}=\frac{1}{2}k{x}^2$ - Total Final Energy: $E_f=0-Mgx+\frac{1}{2}k{x}^2$

Now, we bring it all together using our conservation equation:

Let's rearrange this to reveal a beautiful physical truth:

This equation tells a story: The total gravitational potential energy lost by the block ($Mgx$) is entirely converted into the elastic potential energy stored in the spring ($\frac{1}{2}k{x}^2$).

Now, we just need to solve for $x$. Since we are looking for the maximum extension, we know that $x$ is not zero. Therefore, we can safely divide both sides of the equation by $x$:

Multiplying both sides by 2 and dividing by $k$, we isolate $x$:

Equilibrium vs. Maximum Extension

Look at our final result: $x_{max}=\frac{2Mg}{k}$.

Remember the equilibrium position we discussed earlier? $x_{eq}=\frac{Mg}{k}$.

This reveals a fascinating property of this system: When a mass is dropped onto a spring, its maximum extension is exactly twice its equilibrium extension!

The block falls, accelerates until the equilibrium point, and then decelerates for an equal distance until it stops. This symmetry is a hallmark of simple harmonic motion. By understanding the energy dynamics, you not only find the correct answer but also gain a deeper appreciation for the elegant dance between gravity and elasticity.