Net work done for a mass falling onto a spring – Rankers Physics

Potential Energy & Equilibrium: Practice Problem & Solution

A vertical spring with force constant \(k\) is fixed on a table. A ball of mass \(m\) at a height \(h\) above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance \(d\). The net work done in the process is: (2007)
\(mg(h+d) - \frac{1}{2}kd^2\)
\(mg(h-d) - \frac{1}{2}kd^2\)
\(mg(h-d) + \frac{1}{2}kd^2\)
\(mg(h+d) + \frac{1}{2}kd^2\)

Solution Explained:

To solve this problem, we apply the core principles of Potential Energy & Equilibrium. Understanding the underlying formula is key to arriving at the correct answer below:

Concept: Work done by conservative forces. Formula: \(W_g = mg\Delta h\), \(W_s = -\frac{1}{2}kx^2\).

The total vertical distance the mass falls is \(h+d\), so work done by gravity is \(W_g = mg(h+d)\). The spring is compressed by \(d\), so work done by the spring is \(W_s = -\frac{1}{2}kd^2\). The net work done by these forces is \(W_{net} = W_g + W_s = mg(h+d) - \frac{1}{2}kd^2\).

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