Archive for March 1st, 2016

When Kinetic Energy Meets With Opposing Force

Tuesday, March 1st, 2016

    Objects in motion inevitably meet with opposing forces, a theme which I frequently encounter in my work as an engineering expert.   Today we’ll calculate the opposing force 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.

    Let’s revisit the Work-Energy Theorem formula, whose terms were explained in last week’s blog,

F × d = – ½ × m × v12          (1)

    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 F, the amount of force acting in opposition to the mug’s mass m as it plows a depth d into the litter, we’ll isolate it on one side of the equation, as shown here,

F = [- ½ × m × v12 ] ÷ d          (2)

    So how do we solve for F when we don’t know the value of v1, the mug’s freefall velocity at impact?   We’ll use the fact that The Law of Conservation of Energy tells us that all energies 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 energy held within it as it sat upon the shelf prior to falling.  Equation (2) then becomes,

F = [- m × g × h] ÷ d           (3)

where g is the Earth’s acceleration of gravity factor, a constant equal to 9.8 meters/sec2 , and h is the height from which the mug fell.

Kinetic Energy Meets With Opposing Force

Kinetic Energy Meets With Opposing Force


    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 force acting upon it at the time of impact.   It’s time to plug numbers.

    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 force acting upon it is,

F = [- (0.25 kg) × (9.8 meters/sec2) × (2 meters)] ÷ (0.05 meters)

=  – 98 Newtons

    The mug was subjected to -98 Newtons, or about -22 pounds of opposing force when it fell into the litter, that resistance being presented by the litter itself.

    Next time we’ll see what happens when our mug strikes a hard surface that fails to cushion its impact.    Energy is released, but where does it go?

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog