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Power | power | Lifting a fixed | Potential and Kinetic energy | climbing a Hill | strength versus Energy
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Power, through units of watts, is the rate at which occupational is perform or energy is expended. In a mechanically problem, strength is the moment derivative of power or work. In the current context, strength is the instantaneous product of pressure times distance separated by time:

(1) $ displaystyle ns = fracf dt$ Where: p = Power, watts. F = Force, newtons d = Displacement, meters t = Time, secs

Because $ mfracdisplacementtime$ equates to velocity, over there is a more useful, alternate form of the above equation:

(2) $ displaystyle ns = f v$

Where v = velocity, with units of meters every second.

I questioning my reader to notice about equation (2) that, for strength to be expended, there should be both force and velocity, and also the pressure must be used in the direction the the velocity. A static force without velocity go not need power to keep itself, and also velocity (including rotational velocity) without force likewise does not need power.

Again, for power to be expended: Force and also velocity have to both be present, and also The force must be used in the direction the the velocity.

This meaning may it seems to be ~ excessively strict and detailed, but consider a spinning top — it has both pressure (a centripetal pressure at ideal angles come its velocity) and also rotational velocity, yet no strength is forced to keep the velocity (the optimal will only slow down if friction is present). The factor is that the force"s direction is at best angles to the velocity"s direction.


Energy, v units the joules (or watt-seconds), is strength multiplied by time. This way the expenditure of 100 watt for 30 minute represents an power expenditure the (100 x 30 x 60) 180,000 joules.

Remember that power is constantly conserved. If energy is expended lifting a weight, the power must have had actually some prior form (even occasionally as mass). The plot of lifting the fixed represents an instance of power conversion (potential to kinetic energy), and also after the mass has actually been lifted, the energy expended in lifting it remains as (a) gravitational potential energy and (b) a tiny amount of additional mass.

Lifting a Mass

Remember the discussion over — lifting a mass against gravity calls for power, and power applied over time requires energy. The difference in between power and energy is a Calculus idea (energy is the time integral the power), however it"s not necessary to understand much Calculus to grasp this idea. (For those who desire to learn Calculus, click here.)

We have chose to background the vault example"s ten-kilogram mass native the table to a high shelf, a distance of 3 meters. We will compute the required energy first, then discuss power later, for a reason that will become obvious.

For a given mass m, a height h end which the fixed is to be raised, and a gravitational acceleration g, we compute an energy quantity dubbed work, or pressure times distance, through units the joules:

(3) $W = mgh$


m = Mass, kilograms g = Little-g, discussed earlier h = height over i m sorry the fixed is lifted, meter

To elevator a ten-kilogram mass 3 meters versus the pressure of gravity, we should expend this much energy:

(4) $W = mgh = 10 imes 9.80665 imes 3 = 294.1995 extjoules$

Let"s ring this off to 294 joules for the next discussion. Okay, we have the complete energy required, yet how lot power is needed? mental that strength is equal to pressure times velocity (equation (2) above), and energy is strength multiplied through time. This method all the solutions listed below will work:

A power of 294 watt for one second. A strength of 29.4 watts for ten seconds. A power of 2.94 watts for 100 seconds.

Remember this relationship in between power and also energy — a tiny amount of power expended over a lengthy time can achieve the same task together a large amount of power over a brief time. But in all cases, the energy required is the same.

Potential and Kinetic Energy

In basic mechanical systems, energy has 2 primary forms — potential and also kinetic. Potential power is the energy of position or state — a coiled spring, gas under pressure, a publication on a high shelf. Kinetic energy is energy of movement — an arrowhead in flight, a load being lifted, a satellite in orbit. Plenty of mechanical problems turn the end to be descriptions of power conversions — potential come kinetic and earlier again. And also remember — nevertheless of the details, power is constantly conserved — it might be changed in form, yet it is never created or destroyed.

Here is a detailed summary of the over mass-lift difficulty expressed in regards to energy:

Let"s say there is a coiled feather beneath the ten-kilogram mass, and the spring stores mechanical power equal to 294 joules. Prior to time zero, the feather is preserved from releasing its tension, that energy. The spring"s stored potential energy causes the spring"s fixed to rise slightly, according to this special Relativity principle: (5) $m = fracEc^2$


m = mass, kilograms E = Energy, joules

Because of its potential energy of 294 joules, the coiled feather gains this lot mass:

(6) $m = fracEc^2 = frac294299792458^2 = 3.27 imes 10^-15 extkilograms$

This calculation mirrors that the spring has an energy equivalent mass boost of simply over three femtograms, about the mass of five E. Coli bacteria.

At time zero, the feather is released and also expends that 294 joules of energy raising the ten-kilogram mass three meters.

The spring"s potential power becomes kinetic energy, the power of motion.

The time this motion takes doesn"t matter — together explained over it might be a portion of a second, or it might be numerous minutes. The only issue is even if it is a complete mechanical power of 294 joules is expended.

The spring reaches its complete height and also the fixed is deposited on a high shelf.

What had been mechanical stress and anxiety in the spring became kinetic energy throughout the lift, and also what had actually been kinetic power is now gravitational potential energy in the mass sit on the shelf. The massive now has actually the power equivalent mass rise — 3.27 femtograms — the was lost by the spring.

A automobile must climb a grade in ~ a details speed. We will use the techniques described over to determine just how much speech is required to accomplish a stated climb rate. Below are the details:


auto mass m
: 3000 kilograms Hill steep (rise/run): 0.12. Hill angle φ = $ displaystyle tan^-1(rise/run)$ = 6.84° preferred velocity v: 15 m/s (54 kph).

A nomenclature notice. Hill slopes are occasionally expressed as grade percentages, where the worth is same to:

(7) $ displaystyle grade \% = 100 fracriserun$

Using this nomenclature, ours hill slope of 0.12 would be defined as a 12% grade. To convert such a grade right into an angle, compute:

(8) $ displaystyle phi = tan^-1 left(fracgrade100 ight)$.

When climb a grade, we experience less gravitational force than in a upright lift. The force equation for this case is as such the pressure equation $f = m g$ modification by a slope edge of φ:

(9) $ displaystyle f = m g sin(phi)$


f = Force, newtons m = Mass, kilograms g = Little-g, explained earlier φ = Slope angle

Now that we know exactly how to compute the force, we can find the power required to meet the preferred velocity score of 15 m/s. In a previous page we learned that power is equal to pressure times velocity, therefore:

(10) $ displaystyle p = f v = m g sin(phi) v = 3000 imes 9.80665 imes sin(6.84°) imes 15 = 52,578.69 extwatts$ To transform from watt to horsepower we division by 746 (although there are countless other possible values): 52,578.69 watt / 746 = 70.48 horsepower.

We have developed the power forced to sustain a specific velocity while climbing a grade, now we will compute the energy needed to climb a details distance. Let"s speak the slope proceeds for two kilometers — just how much energy is compelled to climb it?

Remember that energy is power multiplied by time. Additionally remember that the same energy can an outcome from a low strength level applied over a lengthy time, or a high strength level used over a brief time. But perhaps an ext important, the energy required doesn"t depend on the details of the hill, the slope or the time duration, only the vertical height over i beg your pardon the massive is lifted. Every we need to recognize is the vertical lift distance and also the mass.

See more: How Many Meters In 10 Ft - How Many Meters Are 10 Feet

We recognize that the vehicle traveled a street d
of 2 kilometers increase a grade having actually an angle φ that 6.84° (refer to number 2 above). Let"s transform this distance into a vertical height h: (11) $ displaystyle h = d sin(phi) = 2000 imes sin(6.84°) = 238.19 extmeters$ currently we can apply equation (11) come compute the occupational (force times distance) energy required. Remember the the auto has a mass m the 3000 kilograms: (12) $ displaystyle W = mgh = 3000 imes 9.80665 imes 238.19$ = 7.01 megajoules
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