The International System of Units ( SI ) uses seven base units to describe seven fundamental quantities that can be measured by scientists:
Symbol Name Base quantity
second ( s ) time
meter ( m ) length
kilogram ( kg ) mass
ampere ( A ) electric current
kelvin ( K ) thermodynamic temperature
mole ( mol ) amount of substance
candela ( cd ) luminous intensity
Various combinations of these quantities can be used to describe other measurable phenomena. For example, an object’s velocity is the distance ( m ) that an object travels over some period of time ( s ):
velocity ( v ) = ( meters / seconds ) = m/s
The joule ( J ) is a derived SI unit that quantifies energy. Energy within a system can be potential energy ( PE ) or kinetic energy ( KE ). Generally speaking, potential energy is the energy that some entity possesses due to its position, and kinetic energy is the energy of motion.
To derive the joule, we begin with the property of mass. Mass is a measure of an object’s tendency to resist being put into motion or have its existing motion changed via interaction with an outside force. This tendency of matter is a measure of the inertia that an object possesses. If an object is to be displaced from some point of reference, energy must be used to perform work upon the object over some distance:
W = Fd
The SI derived unit of force is the Newton ( N ); it is defined as a product of mass and acceleration:
F = ma
Acceleration is a derived unit that describes how an object’s velocity changes over time under the influence of an imbalanced force. When an object accelerates, the meters the object moves each second changes every second:
a = ( v / s ) = ( [ m / s ] / s ) = m/s2
Thus, the units that comprise a force are as follows:
F = ma = ( kg )( m / s2 )
We are now ready for a rudimentary derivation of the joule:
W = Fd
W = ( kg )( m / s2 )( m ) = kg*m2/s2
Thus, 1 joule is the product of 1 kilogram multiplied by the ratio of 1 square meter over 1 second squared. Notice that the latter part of this term is the square of velocity:
v = ( m / s )
v2 = ( m / s )2 = m2/s2
Since velocity describes motion, the average kinetic energy of mass-in-motion may be described with the following derivation:
KE = ½ mv2
The relationship between work, energy, and displacement is a common source of confusion among physics students. If an object is moved away from a location only to be returned later, its net displacement is zero, and no work has been done upon it! A full discussion of why this is the case requires a discussion of conservative and non-conservative forces. Suffice it to say that an object that is displaced within a conservative force field has the ability to regain its initial energy status. A non-conservative force such as friction initiates energy losses that cannot be recovered ( Ex. heat, sound, etc. ) in accordance with the Second Law of Thermodynamics.
The gravitational force of attraction is a conservative force that is quantified by the following expression:
Fg = Gm1m2/r2
Let’s consider m1 to be the mass of the earth, and m2 to be that of an object on the earth’s surface. The mass of the earth is much, much larger than anything on its surface, and its center-to-surface radius is much, much longer than that of objects sitting on its surface. For this reason, the following assumption can be made:
GMe/re2 = a constant value of ( g ) = 9.80 m/s2
Thus, the force of gravity ( weight ) acting upon an object near the earth’s surface is approximated with this expression:
Fg = mg
Does all of this mean that a motionless rock’s position atop a cliff is maintained without effort? Of course not. If a helicopter lifts a large rock from a cliff only to replace it in its exact position later, the rock will resume its potential energy status, but the helicopter will irreversibly lose energy to entropy. However, the rock has a higher potential energy than anything that is situated closer to the earth’s center. Furthermore, the rock’s energy relative to an onlooker underneath is determined by their relative difference in position from the earth’s center:
PE = Fd = mgh = mg( h2 – h1 )
If we designate the upward direction as being positive relative to the earth’s center, an object on the earth’s surface is located a distance ( h1 ) from the center, and any object situated above it is some distance ( h2 ) from the earth’s center. Under these circumstances, ( h2 – h1 ) yields a positive value, because ( h2 > h1 ). Conversely, an object located beneath the earth’s surface will have an ( h2 ) value that is less than ( h1 ), and its potential energy relative to the earth’s surface will be negative.
How must we quantify the kinetic energy of an object that has fallen from an arbitrary height to some lower position on the earth’s surface? Such an object is under the influence of an unbalanced force up to the moment of impact. Under such conditions, it will not move with a constant velocity. How does the kinetic energy equation apply in this circumstance? As the object loses potential energy, it gains proportional amounts of kinetic energy. This change in energy status is quantified with the following equation:
½ mv2 = mgh
When ( h ) is positive, a net gain in kinetic energy will occur as the object moves from ( h2 ) down to ( h1 ). If ( h ) is negative ( h2 < h1 ), an object thrown upward will lose kinetic energy as it moves to a higher location:
Q: A 1.0 kg mass moves upward with an initial velocity of 5.0 m/s. What is the final velocity of the mass if it moves upward a distance of 1.0 meter? Assume the upward direction to be positive.
A: The initial kinetic energy of the mass can be determined by using the kinetic energy equation:
½ mv2 = ½ ( 1.0 kg )( 5.0 m/s )2 = ½ ( 1 kg )( 25.0 m2/s2 ) = 12.5 J
The loss in kinetic energy is proportional to the upward displacement of the mass. Additionally, since the upward direction is positive, the gravitational acceleration constant ( g ) will have a negative value:
ΔKE = mgh = ( 1.0 kg )( – 9.80 m/s2 )( 1.0 m ) = – 9.8 J
Furthermore,
ΔKE = KEf – KEi
– 9.8 J = KEf – 12.5 J
KEf = 2.7 J
Let’s now determine the final velocity of the mass:
½ mv2 = 2.7 J
mv2 = ( 2 )( 2.7 J ) = 5.4 J
v2 = ( 5.4 J / 1 kg )
v = 2.3 m/s
Let’s now determine what loss of kinetic energy has occurred via usage of the appropriate kinematic equation:
( vf )2 = ( vi )2 + 2a( xf – xi )
( vf )2 = ( 5.0 m/s )2 + 2g( xf – xi )
Note: With the upward direction designated as positive, the gravitational constant of acceleration ( g ) acting in the opposite direction has a negative value.
( vf )2 = ( 5.0 m/s )2 – 2g( 1.0 m )
( vf )2 = ( 25.0 m2/s2 ) – 2g( 1.0 m )
( vf )2 = 25.0 m2/s2 – 19.6 m2/s2
( vf )2 = 5.4 m2/s2
vf = 2.3 m/s
ΔKE = ½ m( v2f – v2i ) = ½ m[ ( 2.3 m/s )2 – ( 5.0 m/s )2i ]
ΔKE = ½ m( 5.3 m2/s2 – 25.0 m2/s2 )
ΔKE = ½ m( – 19.7 m2/s2 )
ΔKE = ½ ( 1 kg )( – 19.7 m2/s2 )
ΔKE = – 9.9 J
RENAISSANCE PHYSICS 