When a spiral spring is compressed by an external force of 200 N, it stores 0.16 J of energy. What amount of energy will it store when compressed by an exte...

Answer Details

For a spring obeying Hooke's law, the elastic potential energy stored is:

\[ E = \frac{1}{2}kx^2 \]

where \( k \) is the spring constant and \( x \) is the compression (or extension). Since the applied force \( F = kx \), we can write \( x = \frac{F}{k} \), and substituting:

\[ E = \frac{1}{2}k\left(\frac{F}{k}\right)^2 = \frac{F^2}{2k} \]

This shows that the energy stored is proportional to the square of the applied force: \( E \propto F^2 \).

For two different forces applied to the same spring:

\[ \frac{E_2}{E_1} = \left(\frac{F_2}{F_1}\right)^2 \]

Substituting the given values:

\[ \frac{E_2}{0.16} = \left(\frac{700}{200}\right)^2 = (3.5)^2 = 12.25 \]

\[ E_2 = 0.16 \times 12.25 = 1.96 \text{ J} \]

The spring stores 1.96 J of energy when compressed by 700 N.

The critical insight is that energy depends on the square of the force, not linearly. Tripling the force does not triple the energy - it increases it by a factor of nine.