Question

1. a 30 kg crate is sliding along with a speed of 5 m/s. if it hits a rough patch of floor that stops the crate over a distance of 1.5 meters, how much frictional force does the floor apply to the crate?

Explanation:

Step1: Use the work - energy theorem

The work - energy theorem states that \(W=\Delta K\). The initial kinetic energy \(K_{i}=\frac{1}{2}mv^{2}\), and the final kinetic energy \(K_{f} = 0\) (since the crate stops). The work done by the frictional force \(W=-F_{f}d\) (negative because the force is opposite to the displacement). So, \(-F_{f}d=\frac{1}{2}mv^{2}-0\).

Step2: Solve for the frictional force \(F_{f}\)

We can rewrite the equation from Step 1 as \(F_{f}=-\frac{mv^{2}}{2d}\). Given \(m = 30\space kg\), \(v = 5\space m/s\), and \(d=1.5\space m\). Substitute the values: \(F_{f}=-\frac{30\times(5)^{2}}{2\times1.5}\).
First, calculate \(30\times(5)^{2}=30\times25 = 750\). Then \(2\times1.5 = 3\). So \(F_{f}=-\frac{750}{3}=- 250\space N\). The magnitude of the frictional force is \(F_{f}=250\space N\).

Answer:

The frictional force applied by the floor to the crate is \(250\space N\).