Children sled down a 41 -m-long hill inclined at \(25^{\circ} .\) At the bottom, the slope levels out. If the coefficient of friction is \(0.12,\) how far do the children slide on the level ground?

First, calculate the initial speed of the children at the bottom of the hill using energy principles. For this, we can use the principle of conservation of energy, which states that potential energy at the top of the slide will be equal to the kinetic energy at the bottom, minus any energy losses due to work done by friction. The potential energy is given by \( m \cdot g \cdot h \) where \( m \) is the mass (which we don't need to calculate as it gets canceled out), \( g \) is acceleration due to gravity, and \( h \) is the vertical height. The vertical height can be calculated from the length of the hill \( s \) and the angle of inclination \( \theta \) using trigonometry: \( h = s \cdot \sin(\theta) \). The work done by friction is defined as \( W = \mu \cdot m \cdot g \cdot d \), where \( \mu \) is the coefficient of friction and \( d \) is the distance (here it's the length of the hill). Equating potential energy and work done by friction, we can solve for the speed \( v \) at the bottom.

Next, calculate how far the children will slide on the level ground before coming to a stop. The children will slide until the kinetic energy at the bottom of the hill is completely converted to work done by friction on the level ground. The work done by friction is defined as \( W = \mu \cdot m \cdot g \cdot d \), and the kinetic energy at the bottom of the hill is \( \frac{1}{2} \cdot m \cdot v^2 \). Equating these two, we solve for the distance \( d \) along the level ground.

Solve the equations from step 1 and step 2 with the given values. Important to note is that the mass \( m \) will cancel out during the calculations. Once all the calculations are completed, simplify the answer to get the final distance the children will slide along the level ground after reaching the base of the hill.