How much work is needed to tip over a stone cube of mass m =…

How much work is needed to tip over a stone cube of mass $m = 100\ \mathrm{kg}$ and edge $a = 0.5\ \mathrm{m}$ about one of its edges?

Explanation of Method

Work to tip over a stone cube.

Step 1: Rise of the centre of mass

The cube tips over once its centre of mass is directly above the edge, i.e. when it stands on the edge with a diagonal of its face vertical. The centre of mass rises from $a/2$ to half the face diagonal:

\[ \Delta h = \frac{a\sqrt2}{2} - \frac{a}{2} = \frac{a(\sqrt2 - 1)}{2} \approx 0.1036\ \mathrm{m}. \]

Step 2: Work

\[ A = mg\,\Delta h = 981 \cdot 0.1036. \]

\[A \approx 102\ \mathrm{J}\]

$\text{standing}$ $\text{balanced on an edge}$ $\tfrac{a}{2}$ $\Delta h$ $A = mg\,\Delta h \approx 101.6\ \mathrm{J}$

To tip the cube, its centre of mass (red) must be raised from $a/2$ to $a\sqrt2/2$, the moment it stands on the edge with the diagonal vertical. After that it falls over by itself.

Result

\[\boxed{\,A \approx 102\ \mathrm{J}\,}\]

Remark

Check. The original result (101.5 J) is correct to rounding (101.6 J). This is the minimum work, done slowly; beyond the balance point the cube falls over by itself.