## Posts Tagged ‘energy’

Monday, June 6th, 2016
As an engineering expert, I often use the fact that *formulas* share a single common factor in order to set them equal to each other, which enables me to solve for a variable contained within one of them. Using this approach we’ll *calculate* the *velocity,* 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 *combining* two equations which each solve for *kinetic energy *in their own way.
Last time we used this *formula* to *calculate* the kinetic energy, *KE*, contained within the piece,
*KE = F*_{F} ×* d* (1)
and we found that it stopped its movement across the floor when it had traveled a distance, *d*, of 2 meters.
We also solved for the frictional force, *F*_{F}, which hampered its free travel, and found that quantity to be 0.35 kilogram-meters/second^{2}. Thus the *kinetic energy* contained within that piece was *calculated* to be 0.70 kilogram-meters^{2}/second^{2}.
Now we’ll put a second equation into play. It, too, provides a way to solve for *kinetic energy,* but using different variables. It’s the version of the formula that contains the variable we seek to *calculate,* *v,* for *velocity*. If you’ll recall from a previous blog, that equation is,
*KE = ½ ×** m ×** v*^{2} (2)
Of the variables present in this *formula,* we know the mass, *m,* of the piece is equal to 0.09 kilograms. Knowing this quantity and the value derived for *KE *from *formula* (1), we’ll substitute known values into *formula *(2) and solve for *v*, the *velocity, *or traveling speed, of the piece at the beginning of its slide.
__Combining Kinetic Energy Formulas to Calculate Velocity__
The ceramic piece’s *velocity* is thus *calculated* to be,
*KE = ½ ×** m ×** v*^{2}
0.70 *kilogram-meters*^{2}/second^{2}= ½ × (0.09 *kilograms*) *×** v*^{2}
now we’ll use algebra to rearrange things and isolate *v* to solve for it,
*v*^{2} = 2 *×* (0.70 *kilogram-meters*^{2}/second^{2}) ÷ (0.09 *kilograms*)
*v* = 3.94 *meters/second =*12.92* feet/second = *8.81* miles per hour*
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
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Tags: distance, energy, engineering expert, friction, frictional force, kinetic energy, mass, velocity, work

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Wednesday, May 25th, 2016
My activities as an **engineering expert** often involve creative problem solving of the sort we did in last week’s blog when we explored the interplay between work and kinetic energy. We used the Work-Energy Theorem to mathematically relate the kinetic energy in a piece of ceramic to the work performed by the friction that’s produced when it skids across a concrete floor. A new formula was derived which enables us to *calculate the kinetic energy* contained within the piece at the start of its slide *by means of the work of friction.* We’ll crunch numbers today to determine that quantity.
The formula we derived last time and that we’ll be working with today is,
**Calculating Kinetic Energy By Means of the Work of Friction**
where, *KE* is the ceramic piece’s *kinetic energy,* *F*_{F} is the frictional force opposing its movement across the floor, and *d* is the distance it travels before *friction *between it and the less than glass-smooth floor brings it to a stop.
The numbers we’ll need to work the equation have been derived in previous blogs. We calculated the *frictional* force, *F*_{F,} acting against a ceramic piece weighing 0.09 kilograms to be 0.35 kilogram-meters/second^{2} and the measured distance, *d,* it travels across the floor to be equal to 2 meters. Plugging in these values, we derive the following working equation,
*KE = *0.35 *kilogram-meters/second*^{2} ×* *2* meters*
*KE = *0.70 *kilogram-meters*^{2}/second^{2}
The *kinetic energy* contained within that broken bit of ceramic is just about what it takes to light a 1 watt flashlight bulb for almost one second!
Now that we’ve determined this quantity, other energy quantities can also be calculated, like the velocity of the ceramic piece when it began its slide. We’ll do that next time.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: distance, electrical energy, energy, engineering expert, frictional force, kinetic energy, mass, velocity, Watt, work, work of friction, work-energy theorem

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Thursday, May 12th, 2016
We’ve been discussing the different forms energy takes, delving deeply into de Coriolis’ claim that *energy* doesn’t ever die or disappear, it simply changes forms depending on the tasks it’s performing. Today we’ll combine mathematical formulas to derive an equation specific to our needs, an activity my work as an **engineering expert** frequently requires of me. Our task today is to find a means to calculate the amount of *kinetic energy* contained within a piece of ceramic skidding across a concrete floor. To do so we’ll combine the frictional force and Work-Energy Theorem formulas to observe *the interplay between work and kinetic energy*.
As we learned studying the math behind the Work-Energy Theorem, it takes work to slow a moving object. Work is present in our example due to the friction that’s created when the broken piece moves across the floor. The formula to calculate the amount of *work* being performed in this situation is written as,
*W = F*_{F} ×*d* (1)
where, *d* is the distance the piece travels before it stops, and *F*_{F} is the frictional force that stops it.
We established last time that our ceramic piece has a mass of 0.09 *kilograms *and the friction created between it and the floor was calculated to be 0.35 *kilogram-meters/second*^{2}. We’ll use this information to calculate the amount of kinetic energy it contains. Here again is the *kinetic energy *formula, as presented previously,
*KE = ½ ×** m ×** v*^{2} (2)
where *m* represents the broken piece’s mass and *v* its velocity when it first begins to move across the floor.
__The Interplay of Work and Kinetic Energy__
The Work-Energy Theorem states that the *work,* *W*, required to stop the piece’s travel is equal to its *kinetic energy,* *KE,* while in motion. This relationship is expressed as,
*KE = W* (3)
Substituting terms from equation (1) into equation (3), we derive a formula that allows us to calculate the *kinetic energy* of our broken piece if we know the frictional force, *F*_{F}, acting upon it which causes it to stop within a distance, *d*,
*KE = F*_{F} ×* d*
Next time we’ll use this newly derived formula, and the value we found for *F*_{F} in our previous article, to crunch numbers and calculate the exact amount of *kinetic energy* contained with our ceramic piece.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: distance, energy, engineering expert, frictional force, kinetic energy, mass, velocity, work, work-energy theorem

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Thursday, April 14th, 2016
Last time we introduced the force of friction, another force in our ongoing discussion about changing forms of energy, and we learned that it’s often a counterproductive force which design engineers and engineering experts such as myself must work to minimize in order to optimize functionality of devices we’re designing. Today we’ll introduce the frictional force formula, which computes the amount of friction present when two surfaces meet.
To demonstrate frictional force, we’ve been working with the example of a shattered mug’s broken ceramic pieces and watching their progress as they slide across a concrete floor. They eventually come to a stop not too far from the point where the mug shattered, because friction causes them to stop. The mass of the ceramic pieces in combination with the downward pull of gravity causes the broken bits to “bear down” on the floor, thereby maximizing contact and creating friction.
At first glance the floor and mugs’ surfaces may appear slippery smooth, but when viewed under magnification we see that both actually contain many peaks and valleys. The peaks of one surface project into the valleys of the other and it’s fight, fight, fight for the ceramic pieces to continue their progress across the floor. The strength of the *frictional force* acting upon the pieces is a factor of their individual weights coupled with the roughness of the two surfaces coming into contact — the shattered pieces and the floor. If friction didn’t exist and no other impediments were in the way, the pieces might travel to the next state before stopping!
__Frictional Force Resists Motion__
Last time we introduced Charles-Augustin de Coulomb, a scientist whose work with friction led to the later development of a formula to calculate it. It’s presented here, and frictional force is denoted as *F*_{F},
*F*_{F} = μ* ×** m ×** g*
where, *m* is the mass of an object making contact with another surface and *g* is the gravitational acceleration constant, which is due to the force of Earth’s gravity. The Greek letter *μ**, *pronounced “mew,” represents the *coefficient of friction*, a number. Numerical values for *μ* were determined by laboratory testing and are recorded in engineering books for many combinations of materials, including rubber on concrete, leather on steel, wood on aluminum, and our own example of ceramic on concrete.
Next time we’ll plug the numbers that apply to our ceramic-on-concrete example into the friction formula and calculate the frictional force at play.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: Charles-Augustin de Coulomb, coefficient of friction, Earth's gravity, energy, engineering expert, friction, friction force formula, frictional force, gravitational acceleration constant, mass, mu

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Tuesday, March 15th, 2016
Last time we watched as the kinetic energy of our falling coffee mug was transformed into the work of creating a crater in a pan of soft kitty litter. Shock absorbing materials are often placed strategically to cushion valuable objects should they fall, and as an **engineering expert** I’ve sometimes had to implement break-its-fall solutions. Today we’ll place our mug into a less kind scenario, one in which it makes impact with the unforgiving *hardness* of a *concrete* floor. In so doing we’ll compare the mug’s *ceramic *to the floor’s *concrete,* and we’ll familiarize ourselves with the* Mohs Scale of Hardness.*
__The Mohs Scale of Hardness, Ceramic vs. Concrete__
Material *hardness* is commonly measured by the* Mohs Scale of Hardness,* which ranks the relative *hardness *of a material by observing how resistant it is to scratching by other materials *harder* than itself. This standard was developed by German mineralogist Friedrich *Moh *in 1812, and it rates objects’ *hardness* on a scale from 1.0, very soft, to 10.0, very hard. A fingernail, for example, ranks 2.5 on the scale, while a diamond ranks 10.0.
Now let’s take a look at the materials in our scenario, a *ceramic *mug and *concrete* floor, and see how they compare. The mug’s *ceramic* was created by mixing together clay, water, and other materials and then heating them in a kiln, a process known as *firing*. This firing causes a chemical reaction that bonds the individual materials tightly together, and when it cools it becomes the product we know as** ***ceramic,* a *hard,* brittle solid which registers at about 7.5 on the *Mohs Scale.*
The floor the mug falls to is poured-in-place *cement,* a compound consisting of primarily limestone, clay, pebbles and sand. When these materials are combined with water a chemical bonding takes place and forms the *hard,* stone-like matter we know as *concrete, *which comes in at about 8.0 on the *Mohs Scale.*
Although the mug’s *ceramic* is comparably *hard* to the floor’s *concrete,* its inherent brittleness, along with certain design features, most notably its handle, causes it to be fragile. Anyone broken a coffee mug lately?
As for the *concrete* floor the mug falls onto, it won’t yield to the mug’s freefall kinetic energy and form a crater like the litter did. So where does the mug’s energy go?
According to the Work-Energy Theorem, most of the mug’s kinetic energy is still converted into work, just as it was when it met up with the litter, but because the *concrete *floor is *harder *and thicker than the mug’s thin *ceramic,* the mug’s kinetic energy at impact falls back on itself rather than transferring externally into the *concrete*. The result is a shattered mug and a mess to clean up.
But we haven’t yet accounted for *all* the mug’s energy. We’ll find out what happens to the rest of it next time.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: brittleness, cement, ceramic, concrete, energy, engineering expert, force, freefall kinetic energy, hardness, Mohs Scale of Hardness, shatters, work, work-energy theorem

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Monday, February 8th, 2016
So far we’ve applied the Work-Energy Theorem to a flying object, namely, Santa’s sleigh, and a rolling object, namely, a car braking to avoid hitting a deer. Today we’ll apply the *Theorem* to a *falling object,* that coffee mug we’ve been following through this blog series. We’ll use the *Theorem* to find the force generated on the mug when it falls into a pan of kitty litter. This *falling object* scenario is one I frequently encounter as an **engineering expert**, and it’s something I’ve got to consider when designing objects that must withstand impact forces if they are dropped.
**Applying the Work-Energy Theorem to Falling Objects**
Here’s the Work-Energy Theorem formula again,
*F ×** d =* ½ *×** m ×** *[*v*_{2}^{2} – *v*_{1}^{2}]
where *F* is the force applied to a moving object of mass *m* to get it to change from a velocity of *v*_{1} to *v*_{2} over a distance, *d*.
As we follow our falling mug from its shelf, its mass, *m*, eventually comes into contact with an opposing force, *F,* which will alter its velocity when it hits the floor, or in this case a strategically placed pan of kitty litter. Upon hitting the litter, the force of the mug’s *falling* velocity, or speed, causes the mug to burrow into the litter to a depth of *d*. The mug’s speed the instant before it hits the ground is *v*_{1, }and its final velocity when it comes to a full stop inside the litter is *v*_{2}, or zero.
Inserting these values into the *Theorem,* we get,
*F ×** d =* ½ *×** m ×** *[0 – *v*_{1}^{2}]
*F ×** d =* – ½ *×** m ×** v*_{1}^{2}
The right side of the equation represents the kinetic energy that the mug acquired while in freefall. This energy will be transformed into Gaspard Gustave de Coriolis’ definition of work, which produces a depression in the litter due to the force of the plummeting mug. *Work *is represented on the left side of the equal sign.
Now a problem arises with using the equation if we’re unable to measure the mug’s initial velocity, *v*_{1}. But there’s a way around that, which we’ll discover next time when we put the *Law of Conservation of Energy* to work for us to do just that.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: energy, engineering expert, falling objects, Gaspard-Gustave de Coriolis, impact forces, kinetic energy, law of conservation of energy, work, work-energy theorem

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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.
__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 × *[*v*_{2}^{2} – *v*_{1}^{2}]
where *W* is the work/energy required to speed up the sleigh team’s mass, *m, *from an initial velocity *v*_{1} to a final velocity *v*_{2}. 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 *meters*^{2}/second^{2})
*W =* 2,925,000 *kilogram*^{2} *· meters*^{2} per *second*^{2} = 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
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Tags: calories, energy, engineering expert, Joules, mass, speed, velocity, work, work-energy theorem

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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 *v*_{1} to a final velocity, *v*_{2,} 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.
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 ×** v*^{2} (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 =* *KE*_{2 }– *KE*_{1} (2)
where *W* is the work performed, *KE*_{1} is the moving object’s initial kinetic energy and *KE*_{2} its final kinetic energy after it has slowed or stopped. In cases where the object has come to a complete stop *KE*_{2} 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 =* [½ *×** m ×** v*_{2}^{2} ] – [½ *×** m ×** v*_{1}^{2}] (3)
Factoring out like terms, equation (3) is simplified to,
*W =* ½ *×** m ×** *[*v*_{2}^{2} – *v*_{1}^{2}] (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 ×** *[*v*_{2}^{2} – *v*_{1}^{2}] (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
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Tags: de Coriolis, distance, energy, engineering expert, force, kinetic energy, mass, velocity, wind force, work, work-energy theorem

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Friday, December 18th, 2015
My work as an **engineering expert** sometimes involves computations of *energy *expended, as when I must determine how much *energy* is required to move something. But sometimes the opposite needs to be calculated, that is, how much *energy* is required to stop something already in motion. That’s the subject of today’s discussion, which we’ll approach by way of the *Work-Energy Theorem*.
*The Work-Energy Theorem* states that the *work* required to slow or stop a moving object is equal to the change in *energy* the object experiences while in motion, that is, how its *kinetic energy* is reduced or completely exhausted. Although we don’t know who to attribute the *Theorem* to specifically, we *do* know it’s based on the previous *work* of Gaspard Gustave de Coriolis and James Prescott Joule, whose *work* in turn built upon that of Isaac Newton’s Second Law of Motion.
Consider the example shown here. A ball of mass *m* moves unimpeded through space at a velocity of *v*_{1 }until it is met by an opposing force, *F*. This force acts upon the ball over a travel distance *d*, resulting in the ball’s slowing to a velocity of *v*_{2}.
**The Work – Energy Theorem Illustrated**
Does the illustration make clear the *Work-Energy Theorem *dynamics at play? If not, return for the second part of this blog, where we’ll clarify things by getting into the math behind the action.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: energy, engineering expert witness, Gaspard-Gustave de Coriolis, Isaac Newton's Second Law of Motion, James Prescott Joule, kinetic energy, moving object, work, work-energy theorem

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Thursday, December 10th, 2015
My work as an **engineering expert** has often required that I perform calculations to quantify the *energy* consumed by electric motors and steam turbines, such as when they *work* together at power plants to generate electricity. Today we’ll see how *work* and *energy** ***share an interesting relationship** that is brought out by examining the units by which they are measured.
Last time we used de Coriolis’ formula to compute work to calculate the amount of *work* performed while pushing a loaded wheelbarrow a distance of 3 meters. We found that in order to move the wheelbarrow that distance, a gardener must exert a force equal to 534 *Newton** • meters* of *work*. That relationship is shown here,
*Work = *178 *Newtons* ×* *3 *meters = *534 *Newton** • meters* (1)
**de Coriolis’ Formula to Compute Work**
The *Newton, *as discussed previously in this blog series, is shorthand notation for metric units of force, and we’ll use those units today to demonstrate the special relationship between *work* and *energy*.
We’ll start by supposing that you’re unfamiliar with the Newton as a unit of measurement. In that case you’d have to employ longhand notation to quantify things, which means you’d be measuring units of force in terms of *kilogram • meters per second*^{2}.
Putting equation (1) in longhand notation terms, we arrive at,
*Work = *178 *kilogram • meters per second*^{2} ×* *3 *meters* (2)
*Work = *534 *kilogram • meters*^{2} per second^{2} (3)
If you’ve been following along in this blog series, you’ll recognize that the unit of measurement used to compute *work, *namely, *kilogram • meters*^{2} per second^{2}, is the same as was used previously to measure *energy. * That unit is the *Joule, *which is considerably less wordy.
Equations (2) and (3) bear out the *interesting relationship* between *work* and *energy — *they share the same unit of measure. This relationship would not be apparent if we only considered the units for *work* presented in equation (1).
So following standard engineering convention where *work* and *energy* are expressed in the same units, the *work* required to push the wheelbarrow is expressed as,
*Work = *534 *Joules*
Yes, *work* and *energy* are measured by the same unit, the *Joule*. But, *energy* isn’t the same as *work*. *Energy* is distinguished from *work* in that it’s the *measure of the ability to perform ***work**. Stated another way, *work* cannot be performed unless there is *energy* available to do it, just as when you eat it provides more than mere pleasure, it provides your body with the *energy* required to perform the *work *of pushing a wheelbarrow through the garden.
Next time we’ll see how *work* factors into the *Work Energy Theorem*, which mathematically relates *work* to *energy*.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: de Coriolis' formula to compute work, electric motors, energy, engineering expert, generate electricity, joule, Newton meters, Newtons, power plants, steam turbines, unit of energy, units of force, work, work energy relationship, work required

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