The crate is moving at a constant speed, so the net force acting on it is zero. The pulling force is balanced by the frictional force.

The frictional force \(F\) is given by \(F = \mu R\), where \(R\) is the normal reaction force.

Since the crate is on horizontal ground, \(R = mg\), where \(m = 500 \text{ kg}\) and \(g = 9.8 \text{ m/s}^2\).

Thus, \(R = 500 \times 9.8 = 4900 \text{ N}\).

The pulling force is 2500 N, so \(2500 = \mu \times 4900\).

Solving for \(\mu\), we get \(\mu = \frac{2500}{4900} = 0.5\).