As an engineering expert, I often use the fact that the calculate or speed, at which the broken bit of ceramic from the coffee mug we’ve been following slides across the floor until it’s finally brought to a stop by friction between it and the floor. We’ll do so by velocity, two equations which each solve for combiningin their own way.kinetic energy Last time we used this the kinetic energy, calculateKE, contained within the piece,
d (1)and we found that it stopped its movement across the floor when it had traveled a distance, We also solved for the frictional force, ^{2}. Thus the contained within that piece was kinetic energy to be 0.70 kilogram-meterscalculated^{2}/second^{2}. Now we’ll put a second equation into play. It, too, provides a way to solve for calculate,v, for . If you’ll recall from a previous blog, that equation is,velocity
Of the variables present in this m, of the piece is equal to 0.09 kilograms. Knowing this quantity and the value derived for KE from (1), we’ll substitute known values into formula(2) and solve for formula v, the or traveling speed, of the piece at the beginning of its slide.velocity,
The ceramic piece’s to be,calculated
0.70 kilograms) × v^{2}now we’ll use algebra to rearrange things and isolate
× (0.70 kilogram-meters) ÷ (0.09 ^{2}/second^{2}kilograms)
Our mug piece therefore began its slide across the floor at about the speed of an experienced jogger. This ends our series on the interrelationship of energy and work. Next time we’ll begin a new topic, namely, how pulleys make the work of lifting objects and driving machines easier. Copyright 2016 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________ |

Tags: distance, energy, engineering expert, friction, frictional force, kinetic energy, mass, velocity, work