Archive for January, 2016

Applying the Work-Energy Theorem to Braking Distance

Friday, January 29th, 2016

    I’m sometimes called upon to render an engineering expert opinion on auto accidents, and in our last blog we stretched this application to a scenario in which Santa’s sleigh collided with the opposing force of a strong wind.   At that time we used the Work-Energy Theorem to calculate the amount of food energy Rudolph and his team required to regain speed and get back on schedule.   Today we’ll use the Theorem to analyze the forces at play in another deer scenario and calculate the braking distance a car needs to avoid hitting one on the highway.

    The average sedan has a mass of about 1,500 pounds, or 680 kilograms.   In our example it’s driving down the highway at a speed, or velocity, of 30 miles per hour, which equates to it covering a distance of 13.3 meters, or just under 44 feet, per second.

    A deer jumps onto the highway, 60 meters in front of the car.   The alert driver slams on the brakes, which exert 1200 Newtons of stopping force on the car.   If you’ll recall from past blogs in this series, the Newton is the metric unit used to measure force.


 What Is Safe Braking Distance?What is Safe Braking Distance?


    Did Bambi survive?   Let’s use the Work-Energy Theorem to find out.   Here it is again,

F × d = ½ × m × [v22v12]

where, F is the braking force used to slow a car of mass m, from an initial velocity of v1 to a final velocity of v2 in a braking distance, d.   The car will eventually come to a complete stop as the driver attempts to avoid hitting the deer, so its final velocity, v2, will be zero.   The Work-Energy Theorem is most often stated in terms of metric units, the measuring unit of choice in the scientific community, and we’ll follow suit with our math.

    Inserting these values into the equation, we get,

[1200 Newtons] × d

 = ½ × [680 kilograms] × [(0)2 – (13.3 meters per second)2]

    Using algebra to solve for d, the braking distance, we arrive at,

d = ½ × [680 kilograms] ×  [(0)2 – (13.3 meters per second)2] ÷  [1200 Newtons]

d = 50.11 meters

    The car stopped 50.11 meters from the point when the driver slammed on his brakes, just about 10 meters short of hitting the deer.   Bambi lives to leap another day!

    Next time we’ll use the Work-Energy Theorem to assess the fate of the falling coffee mug we introduced in past blogs when we opened our discussion on the different forms of energy.


Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



Work and Energy, Speed, and Calories

Wednesday, January 20th, 2016

    In my work as an engineering expert I’ve never had to convert Joules of work-energy into calories, but that’s exactly what we’ll be doing together today.   Last time we applied the Work-Energy Theorem to the progress of Santa’s sleigh and found that an opposing wind force of 3848.7 Newtons  –or 865.2 pounds for those of us who are American–  slowed his team from an initial velocity of 90 meters per second to a final velocity of 40 meters per second and that it happened over a distance of 760 meters.   Today we’ll see how many calories the reindeer need to expend to get them back up to full delivery speed.

 Work Energy Theorem and Reindeer

Prancer Loves Oats

    Now we know that Santa successfully made all his deliveries on time this past Christmas, so that means that at some point his reindeer team was able to get back up to full sleigh-flying speed.   They did it by expending extra energy.   We’ll use the Work-Energy Theorem to find out how much energy that equates to.  Here’s the Theorem again,

W = ½ × m × [v22v12]

where W is the work/energy required to speed up the sleigh team’s mass, m, from an initial velocity v1 to a final velocity v2.   For a refresher on the special relationship between  work and energy, see our past blog on the subject.

    If Santa’s sleigh has a mass of 900 kilograms and its speed must increase from 40 to 90 meters per second, then the work required to do so is calculated as,

W = ½ × (900 kilograms) × [(90 meters/second)2 – (40 meters/second)2]

W = ½ × (900 kilograms) × (6,500 meters2/second2)

W =  2,925,000 kilogram2 · meters2 per second2 = 2,925,000 Joules

    So Rudolph and his buddies had to expend 2,925,000 Joules of energy to perform 2,925,000 Joules of work.   To understand where Rudolph and his team got that energy, we must state things in terms of nutritional value, that is, units of calories.

    Did you know that 1 calorie is equal to 4,184.43 Joules?   Applying that equivalency to our situation we get,

Nutritional Energy Required = (2,925,000 Joules) × (4,184.43 Joules/calorie)

= 699.02 calories

    The net result is Santa’s team expended a total of 699.02 calories for all the reindeer to regain their full speed of 90 meters per second.   That’s the nutritional energy found in slightly more than one cup of oats.   Now everybody knows that Santa takes good care of his reindeer, so undoubtedly they were fed plenty of oats and hay before takeoff.   This was stored in their body fat for future, on-demand use.

    Sadly, Christmas is over, and it’s time to get back to the more mundane aspects of life.   Next time we’ll apply the principles behind the Work-Energy Theorem to calculate the braking force required to stop a car in motion.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



The Work-Energy Theorem Applied to Santa’s Sleigh

Monday, January 11th, 2016

    Last time my engineering expertise was put to the test when it was discovered that Santa’s sleigh was being hampered by a strong gust of wind.   At that time we introduced the Work-Energy Theorem to determine how strong the wind’s opposing force was, and today we’ll work with the Theorem to compute just what Rudolph was up against.   Here again is the expanded, workable version of the Work-Energy Theorem as introduced last time,

F × d = ½ × m × [v22 – v12]

where F is a force acting upon a moving object of mass m over a distance d to slow it from an initial velocity of v1 to a final velocity of v2.

    Applying the Theorem to the dynamics at play in Santa’s situation, F is the opposing wind force, which acts over a distance, d, to slow his sleigh from an initial velocity of v1 to a final velocity, v2, thus presenting Rudolph and his buddies with a real delivery challenge.

Rudolph Struggles Against a Fierce WindRudolph Struggles Against a Fierce Wind 



    If we know that Santa, his sleigh and reindeer have a combined mass of 900 kilograms — which is pretty standard for a fully loaded sleigh and reindeer team — and their initial velocity was 90 meters per second, final velocity 40 meters per second, and the distance over which the slowing took place was 760 meters, then the formula to calculate the opposing wind force becomes,

F × d = ½ × m × [v22 – v12]

F = ½ × m × [v22 – v12÷ d

F = ½ × (900kg) × [(40 meters/sec)2(90 meters/sec)2] ÷ 760 meters

F = -3848.7 Newtons = -865.2 Pounds

    The minus sign signifies that the wind must exert an opposing force of 865.2 pounds in order to slow Santa’s sleigh down.

    In order for Santa to get back on his delivery schedule, Rudolph is going to have to make up for lost time by expending extra energy.   We’ll see how he does that next time.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



The Math Behind the Work-Energy Theorem

Friday, January 1st, 2016

    As an engineering expert I’ve applied the Work-Energy Theorem to diverse situations, but none as unique as its most recent application, the progress of Santa’s sleigh.   Last week we saw how Santa and his reindeer team encountered a wind gust which generated enough force to slow them from an initial velocity of v1 to a final velocity, v2, over a distance, d.   Today we’ll begin using the Work-Energy Theorem to see if Santa was able to keep to his Christmas delivery schedule and get all the good boys and girls their gifts in time.

Santa and the Work Energy Theorem

    Before we can work with the Work-Energy Theorem, we must first revisit the formula it’s predicated upon, de Coriolis’ formula for kinetic energy,

KE = ½ × m × v2                            (1)

where, KE is kinetic energy, m is the moving object’s mass, and v its velocity.

    The equation behind the Work-Energy Theorem is,

W = KE2 KE1                                    (2)

where W is the work performed, KE1 is the moving object’s initial kinetic energy and KE2 its final kinetic energy after it has slowed or stopped.   In cases where the object has come to a complete stop KE2 is equal to zero, since the velocity of a stationary object is zero.

    In order to work with equation (2) we must first expand it into a more useful format that quantifies an object’s mass and initial and final velocities.   We’ll do that by substituting equation (1) into equation (2).   The result of that term substitution is,

W =× m × v22 ] – [½ × m × v12]      (3)

    Factoring out like terms, equation (3) is simplified to,

W = ½ × m × [v22v12]                        (4)

    Now according to de Coriolis, work is equal to force, F, times distance, d.   So substituting these terms for W in equation (4), the expanded version of the Work-Energy Theorem becomes,

F × d = ½ × m × [v22v12]                 (5)

    Next time we’ll apply equation (5) to Santa’s delivery flight to calculate the strength of that gust of wind slowing him down.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog