An object with mass m = 2.6 kg is dropped from rest at height h above a massless spring with spring constant k = 494.3 N/m, which is at its equilibrium position. The block falls onto the spring, compressing it by a distance d = 0.66 m from its equilibrium position before momentarily coming to rest. Calculate the height h from which the block was dropped.

To calculate the height from which the object was dropped, we'll use conservation of energy. The energy is conserved between the initial drop point and the instant where the block is momentarily at rest after compressing the spring.

At the beginning, when the object is at height h above the spring, it only has gravitational potential energy, which is given by the formula \( PE_{gravity} = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (\( \approx 9.81\ \text{m/s}^2 \)), and \( h \) is the height.

When the object is compressed at the bottom of the spring, it has spring potential energy but no gravitational potential energy (assuming h is measured from the point where the block just touches the spring which is in equilibrium). The spring potential energy is given by \( PE_{spring} = \frac{1}{2}kd^2 \), where \( k \) is the spring constant and \( d \) is the compression.

Setting the gravitational potential energy equal to the spring potential energy, we can solve for the height h:

\[ mgh = \frac{1}{2}kd^2 \]

Rearranging for \( h \), we get:

\[ h = \frac{\frac{1}{2}kd^2}{mg} \]

Now we'll plug in the values:

\[ h = \frac{\frac{1}{2}(494.3\ \text{N/m})(0.66\ \text{m})^2}{(2.6\ \text{kg})(9.81\ \text{m/s}^2)} \]

\[ h = \frac{\frac{1}{2}(494.3)(0.4356)}{(2.6)(9.81)} \]

\[ h = \frac{107.93478}{25.506} \]

\[ h \approx 4.233\ \text{m} \]

Therefore, the height h from which the block was dropped is approximately 4.233 meters.