Juana slides a crate along the floor of the moving van. The coefficient of kinetic friction between the crate and the van floor is \(0.120 .\) The crate has a mass of \(56.8 \mathrm{kg}\) and Juana pushes with a horizontal force of \(124 \mathrm{N}\). If \(74.4 \mathrm{J}\) of total work are done on the crate, how far along the van floor does it move?

To find the kinetic friction force between the crate and the van floor, we can use the formula \(F_{friction} = \mu * F_{normal}\), where \(\mu\) is the coefficient of kinetic friction and \(F_{normal}\) is the normal force. Since the crate is not moving vertically, the normal force is equal to its weight (\(mg\)), where \(m\) is the mass and \(g\) is the gravitational acceleration (\(9.81 \, \text{m/s}^2\)). So, we have: \(F_{friction} = \mu * F_{normal} = 0.120 * (56.8 \, \text{kg} * 9.81 \, \text{m/s}^2)\) Calculating this expression gives us: \(F_{friction} = 66.7056 \, \text{N}\)

To calculate the work done by friction, we will use the formula \(W = F * d * \cos \theta\), where \(W\) is the work done, \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and displacement vectors. Since the friction force and the displacement are in opposite directions, the angle \(\theta\) between them is \(180^{\circ}\), which gives us \(\cos \theta = -1\). We have the equation: \(W_{friction} = F_{friction} * d * \cos \theta\) But we want to find the displacement (\(d\)). We know the total work done (\(W_{total} = 74.4 \, \text{J}\)) and the work done by the applied force (\(W_{applied} = F_{applied} * d * \cos \theta\), where \(\theta = 0\) as the force applied by Juana is horizontal). The net work done is: \(W_{total} = W_{applied} + W_{friction} = F_{applied} * d * \cos 0^{\circ} + F_{friction} * d * \cos 180^{\circ}\) We can rearrange the equation to solve for \(d\): \(d = \frac{W_{total}}{F_{applied} * \cos 0^{\circ} + F_{friction} * \cos 180^{\circ}}\)

Now we can plug in the known values to find the distance \(d\): \(d = \frac{74.4 \, \text{J}}{(124 \, \text{N} * \cos 0^{\circ}) + (66.7056 \, \text{N} * \cos 180^{\circ})}\) Evaluating this expression, we find: \(d = 1.2 \, \text{m}\) So, the crate moves \(1.2\) meters along the floor of the moving van.