Abeka Physics Section 12.1 pg. 172 - 178 Flashcards
Define energy
> energy - capacity to do work
Describe energy
> how much effect an entity (whether made of matter or nonmatter) can have on other entities
energy of something is measured by how much disturbance it can cause, or by its tendency to change another object’s present state through interaction
Describe how to derive equation KE = 1/2 mv^2 from F=ma
> suppose isolated object of mass “m” is accelerated from rest along straight line at constant rate “a” for time “t”
uniform acceleration is due to force “F” in direction of motion
if “F” is net force (sum of all forces) acting on object, Newton’s 2nd law requires that F=ma
work done on object is given by W=Fd=(ma)(d)
in case of uniform acceleration from rest v(sub i) = 0, d= 1/2 at^2
Equation F=ma and d= 1/2 at^2 allow substitution in equation W = Fd = (ma)(d) allowing for
W = (ma)(1/2 at^2) or W = 1/2 m(at)^2
if v(sub i)=0, v=at where v is final velocity due to uniform acceleration from rest. thus, after substitution of
“v” for “at” in equation W= 1/2 m(at)^2, one has
W=1/2 mv^2
this work is done in accelerating any isolated object of mass “m” from rest to final velocity “v”
work done on such object is same as energy it receives, which is entirely energy of motion–that is, kinetic energy (KE).
therefore, kinetic energy can be put in place of “W” in equation W=1/2 mv^2
derivation of this equation assumes that object is isolated and has undergone uniform acceleration from rest
but equation is valid even if these assumptions are not met
in other words, object moving at “v”, regardless of its prior motion or environment, has kinetic energy of 1/2 mv^2
thus equation KE=1/2 mv^2 is called kinetic energy equation
Define kinetic energy
> kinetic energy (KE) - energy of motion
Give the kinetic energy equation
> KE = 1/2 mv^2
Describe expression 1/2 mv^2
> expression 1/2 mv^2 tells how much work is required to accelerate isolated object to final velocity “v” from initial velocity of “v(sub f)”, work done on object is equal to gain in kinetic energy
Describe work-energy theorem
> work-energy theorem is relationship between work and energy
“if same object is accelerated to final velocity of “v(sub f)”, work done on object is equal to gain in kinetic energy”
this equation represents work-energy theorem
W = deltaKE = KE(sub f) = KE(sub i)
substituting equation KE= 1/2 mv^2 causes theorem to read as W = 1/2 mv(sub f)^2 - 1/2 mv(sub i)^2
if v(sub i) is zero, equation
W= 1/2 mv(sub f)^2 - 1/2 mv(sub i)^2 reduces to
W= 1/2 mv^2
where friction is involved, force of friction must be subtracted from applied force in order to obtain net force acting on object
work done by this net by this net force is found in
W= 1/2 mv^2 and W= 1/2 mv(sub f)^2 - 1/2 mv(sub i)^2
since work is shown to be equal to change in kinetic energy, both are measured in same units
is customary to treat kinetic energy as quantity of joules (SI units) or foot-pounds (U.S. customary system)
like work, kinetic energy is scalar quantity: no direction is specified
Solve this: a bullet with a mass of 10.0 g is moving at a speed of 1.00 x 10^3 m/s. A bull with a mass of 4.00 x 10^2 kg is charging at a speed of 5.00 m/s. Which has the greater kinetic energy?
> the kinetic energy of both moving objects is given by equation KE= 1/2 mv^2.
for bullet, m is 10.0 g, or 0.0100 kg, and v is 1.00 x 10^3 m/s.
thus, after substitution, KE= 1/2(0.0100kg)(1.00x10^3m/s)^2
KE= 5.00 x 10^3 kg m^2/s^2 = 5.00 x 10^3 J
since newton is a kg m/s^2, and since joule is Nxm, analysis of above answer allows for simplification of units:
KE = 5.00 x 10^3 J
for bull, m is 4.00 x 10^2 kg and v is -5.00 m/s (traveling in opposite direction).
thus, KE= 1/2(4.00 x 10^2 kg)(-5.0 m/s)^2
KE= 5.00 x 10^3 J
thus, their kinetic energies are the same
Describe relative contributions of mass and velocity to kinetic energy
> bullet in example above is so small that, if it were lying in grass, bull would hardly notice it even if he stepped on it
mass of bull is 40,000 times greater
yet when bullet is fired, it develops as much kinetic energy as bull when he is charging ferociously
reason is that, while kinetic energy depends on first power of mass, kinetic energy also depends on second power of velocity–that is, on velocity squared
therefore, bull and bullet have equal energy if velocity of bullet is only 200 times greater
if kinetic energy is to be maximized, it is advisable to increase velocity at expense of mass wherever possible
for example, golfer doesn’t hit ball with club as big as tree limb
rather, he uses lightweight club that can be swung at high velocity
as result, maximum kinetic energy is developed in club and then imparted to ball
Describe work done to stop object
> in previous chapter, positive work is done when force is applied in direction of motion
negative work is done when force is applied in direction opposite to motion
imagine our charging bull running headlong into rail fence
fence exerts force on bull and causes him to slow down, at least, momentarily; thus, fence does negative work on bull
but because action equals reaction, bull also applies force to fence, causing pieces of wood to accelerate in all directions; thus, bull does positive work on fence
kinetic energy of object can be interpreted in 2 ways: (1) kinetic energy is equal to positive work that 1 object can do on another while coming to a stop v(sub f)=0, and (2)kinetic energy is equal in magnitude to negative work required for one object to be stopped by another
Solve this: a hammer with an initial velocity of 9.2 m/s hits a nail. The hammer weighs 0.50 kg. How much work does it do in driving the nail into a piece of wood?
> assuming that the hammer is slowed to a stop by the nail, the work done by the hammer after hitting the nail is equal to the previous kinetic energy of the hammer. Find W from equation W= 1/2 mv^2, regarding m as 0.50 kg and v as 9.2 m/s.
W= 1/2(0.50kg)
Define potential energy
> “energy object possesses by virtue of its position or configuration”
“energy object acquires when it is exposed to force”
Describe potential energy
> many times defined as “energy object possess by virtue of its position or configuration
depends on position or configuration only because these govern what forces act upon object
general definition “energy object acquires when it is exposed to force”
energy called potential because such object will, if it responds to force, acquire actual energy of motion; but object has potential energy whether or not it is presently free to respond to force
Describe gravitational potential energy
> object in vicinity of earth has gravitational potential energy as result of being under influence of gravity
force needed to raise any object is slightly more than its weight until it reaches desired velocity
force equal to weight is sufficient to keep it moving
for all practical purposes, lifting force can be regarded as equal to weight [F(sub w)]
since work done in raising object to height “h” is found by W = F(sub w) x h
since weight is same as product “mg”, W = mg x h
if potential energy is called “PE”, the PE of object at ground level is arbitrarily assigned value of zero
whereas, PE of object at height “h” is equal to energy object acquires when it is lifted from ground level to this height
this acquired energy is equal to work performed on object against gravity in raising object to height “h”
since work done on object is same as potential energy (PE) it receives, we can also state:
W= delta PE = PE(sub f) - PE(sub i) or,
W= delta PE = mgh(sub f) - mgh(sub i)
if object lifted from ground [h(sub i)] = 0, then
W= delta PE = mgh(sub f) - 0 = mgh(sub f)
Describe gravitational potential energy with example of throwing ball into the air
> if you throw ball up into the air, ball has certain initial velocity and therefore certain kinetic energy when it leaves your hand
this kinetic energy is equal to work accomplished by your throwing motion
as ball rises, its velocity decreases under influence of gravitational force
when ball eventually reaches its maximum height, its final velocity is zero
it may seem that kinetic energy of ball as it left hand has disappeared, but really there has been no loss of energy; instead, kinetic energy has been converted to potential energy
as stated in equation W= PE(sub f) - PE(sub i) or W= mgh(sub f) - mgh(sub i) , potential energy of ball at maximum height is equal to work done on ball as it is lifted against pull of gravity
equation W= mg x h states amount of work as mgh even though, as you throw ball upward, you do not exert constant force of mg over distance h
instead, you apply much stronger force over much shorter distance
nevertheless, product of this force and distance can only be mgh since PE of object depends only on its present position; how it arrived there is irrelevant
energy stored in ball at top of its flight (its potential energy) is reconverted to kinetic energy as ball falls back down into energy at end of flight (neglecting air resistance) is equal to kinetic energy at beginning
