Objects in motion inevitably meet with opposing engineering expert. Today we’ll calculate the opposing our exemplar coffee mug meets when it falls into a pan of kitty litter, thus transforming its freefalling kinetic energy into the work required to move through clay litter.forceLet’s revisit the Work-Energy Theorem formula, whose terms were explained in last week’s blog,
The left side of this equation represents the mug’s work to move through the litter, while the right side represents its kinetic energy, which it gained through freefall. To solve for m as it plows a depth d into the litter, we’ll isolate it on one side of the equation, as shown here,
d (2) So how do we solve for are equal, and we’ll eliminate the part of Equation (2) that contains this unknown variable, that is, the right side of the equation which deals with kinetic energy. In its place we’ll substitute terms which represent the mug’s potential energy, that is, the latent energies held within it as it sat upon the shelf prior to falling. Equation (2) then becomes,energy
where h is the height from which the mug fell.
So if we know the mug’s mass, the distance fallen, and the depth of the crater it made in the litter, we can determine the stopping Let’s say our mug has a mass of 0.25 kg, it falls from a height of 2 meters, and it makes a crater 0.05 meters deep. Then the stopping
× (2 meters)] ÷ (0.05 meters)= – 98 The mug was subjected to -98 Newtons, or about -22 pounds of opposing Next time we’ll see what happens when our mug strikes a hard surface that fails to cushion its impact. Copyright 2016 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________ |