This blog created by Fariel Mohan is to assist MATH110D students in understanding foundational mathematical concepts.
Saturday, March 1, 2008
K.E. = 1/2 m v squared
If you was the scientist to discover the K.E. formula starting from the fact that K.E. is proportional to the velocity to a power by the mass. Illustrate your steps to lead to the above formula.
hmmmmmm. miss u taxin men brain wih this question any how i go try so K.E. proportional to m v^x let us say the power is x K.E. = kmv^x log K.E. = log kmv^x log K.E. = log k + log mv^x log K.E. = log k + log m + log v^x whey ah stick i probably goin wrong someone HELP
moose why it is u bring logs into the equation? u confusing me.which part the 1/2 gone? miss i think there's an error in the post. K.E is proportional to half the velocity to a power by the mass.
wel i would try to help u out...wel actually help myself
ok Ke is porpotional to mv^x then if u introduce an = sign then u need to introduce a constant k... im with u up to here... now my question......how did u get logs??? u just use logs jus so ???
i think i would have used graphs an different set of values in small increments..e.g 0.0000001, 0.000001 ,0.00001 ect..
CAPTAIN VEGETABLE!!! HOW COME U EH POST ANYTING ON THIS ONE AS YET. LIKE YUH SUFFER FROM A NERVOUS BREAKDOWN OH WAT????? HAVE SOME MEAT!!! AHHHHH!!! HAAAA!! HA!! HA HA HA!!!! Anyway back to more important stuff!!! Ha Ha Ha Ha!!! Oh Gosh!!! Enough!!! Enough!! From this question I cud gather that K.E directly prop to mv^p so K.E = kmv^p. Im kinda lost. Any ideas CAPTAIN VEGTABLE!! I KINDA NEED U!!! ARE U DERE???? Moose why u take logs????
now in a scientific experiment there will be the use of a graph for this discovery. and it is only through the means of a graph showing results of experiments carried will this question be solved
oh and spartan if u read the question u would see that the post said that ke is proprtional to THE VELOCITY TO A POWER MULTIPLY BY MASS THERE WAS NO MENTION OF 1/2 OR THE POWER OF 2
well, saying that K.E. is proportional to velocity to a power by a mass would be represented as: K.E. = k(v^x)m but a value must be present to solve for k and then you differentiate, or you don't differentiate, or i probably on the wrong topic, WELL??
dmcxlKE: mv^2/2...KE: mv^2/2.........KE: mv^2/2....KE: mv^2/2...HOW HOW HOW!!! IM CLUELESS!! HELP!!!!!! EVERYBODY TALKING ABOUT LOGS!!! HOW LOGS CUM IN DIS????
i really don't know how to do it in logs but i can derive it in a next way(using motion equations)but we didn't do that in class so ...... can someone help us all with this question?
i have no idea how to derive this... but i can sure use it to find angular velocity in a solid wheel... rasif never showed us how it was derived i think
anyway, this formula is used in physics a lot, so instead of using logs and so on, can't we just use other formulas depicting force or velocity and just substitute parts from them into one formula?
what i mean is... instead of using logs and graphs and all those things, if we use certain elements of simple formulas such as 'f=ma' where 'a = change in v / t' and so on, won't we be able to get back the same formula?
moose y u bring in logs for it looks more intimidating now. i would like som help in clearin up this plz. im not 2 good in physics and bringing it into maths is jus making more trouble for me so can i get som help
first of all what is kinetic energy? kinetic energy is used to describe the amount of energy a moving energy that a body contains. so what about an object that is subjected to free fall. since the acceleration due to gravity is 9.8m/s[2], how can the kinetic energy of the body be calculated.
i think its possible.we may be able to change aroung de formulars...but then if we do that what about de odda people who never did physics ? they wouldnt understand any of it...we would have to go into alot of xplaination..
@ yoshi
what is integral? what does it means? i confuse day papi..so xplain dat concept
fredbugger i think they using logs here coz ent scientist does try all kinda methods to solve problems ? well rite dem wanna be scientists so let them be!!! lol
wel may be.. i for one aint no scientist so i dunno nah..i jus rember dat miss say they does use small changes in the values when plotting graphs...like 0.00001 , 0.0001 and so on...
This is my version for this problem. If I am a scientist. I would most likely have to prove this using a graph with a curve.In order to use the graph to get information I would have to use logs. take logs on both sides
log K.E= -2log m + 2log v log K.E= log m^(-2) + log v^(2) log K.E= log m^(-2)x v^(2) log K.E= log m/2 x v^(2) log K.E= log 1/2 x m x v^(2) drop logs
this question is a bit tricky so i'll try my best and see if i can do this, so if K.E. is proportional to the velocity to a power we have: K.E.= k v^x where k is a constant of proportionality, v is velocity and x is its unknown power. since this is then multiplied by a mass then the formula should be K.E. = kmv^x I know it a bit different from what every one else is saying with all this logs and integral but i think this is the basic formula
But moose why you take logs for??? If that is how it really started I lost that is going to defeat the purpoes of what I ahave in mind. K.e is movement due to the movement of the mass times 1/2 the velocity^2 .
If i was a scientist, i would say well k.e depends on movement and movement depends on speed or velocity. K.E. is directly proportional to v^2 this meaning as velocity increases the energy would also increase, assuming that mass is constant.OK! but linear momentum is also proportional on velocity. so i would think that the power plays a very important role but what is it. if u differentiate K.E. u would get linear momentum, which is mv so we can se there is a relationship. ask urself what is linear momentum? i think momentum is the power residing in a moving object. The thing is mv presents itself graphically as a straight line but 1/2mv^2 is a curve...i will get bak to u on this miss, i hav to go to ur class. i need to brain storm some more
Ok...i tink papa vigil has opened up a door here. i dont know if im correct on this one but he talked about a straight line. Remember momentum is equal to mass by velocity. remember mass is constant. so if we plot a graph of momentum against velocity we wud get a straight line. I tink the gradiant of this line suppose to be 2 and the y intercept suppose to be 1/2. sooooo..um we can now put this in the form ax^n..ENT??? wat u all tink?????
Ke= kmv^x where k is the constant of proportionality, m is the mass and x is the unknown power. taking out the constant we get KE= mv^x i think that the relationship can be calulated experimentally. then write the equations keeping each variable constant then manipulating it to get the factor by which they should be multiplied.
thought about it some more miss, and where i stuck b4 was because i was thinking why would they square v in the first place...why it could not be 1/3mv^3 or 1/4mv^4...but then i realised that when u square v, ur units will be (m/s)^2 which is (m^2/s^2), which can be seen to be m(ms^-2)...do u see that u units would infer that there is a distance by a constant acceleration...then i thought P.E=K.E., so the the P.E a body has, at a height is equal to the K.E it has as it travels throuh that distance...so i would presume as the scientist, that v^2 is the constant acceleration by that distance as mentioned b4..the coefficient of 1/2 is used to get mv when 1/2mv^2 is differentiated
this qusetion has notting to do with logs . first gust dress up the given equation to the required equation. Be very carefull of where u apply your rules.
But before u do any calculations, Remember u are a scientist, and u have to work on the concepts first, so u will plot a graph and fing the constant, or make any diductions that is usefull.
Now u may solve the equation: K.E. = kmv^x Thin may tie u up because so you put:
K.E. = kmv
K.E. = km(v^2)^1/2
u then square root the V and after u square V to get rid of the square root.And u will end up with the answer bellow
miss, i like this question! because it makes u think about concepts and not just practical formulatiion...captain vegetable why did u not post anything 4 this question, is this question too high of a caliber than ur mind's capacity can handle
according to papa vigil and to add what he say but notin the rouhg sense captain veg if u could answer this question please do if you have the knowledge please share it for curious blogbakes like me it will be much appreaciated
oh and captain vegetable i hope yuh belly not thin because it seems no one on the blog likes u but for me u does make me real think so i guess u make sense
In continuation to my previous comment, i would establish that Force= mass x acc. and energy or work = force x dist...so from the equation k.e= 1/2mv^2, u actually hav a mass(m), an acc.(ms^-2), a dist.(in meters).
why do u need CAPTAIN VEGETABLE when the master is here. seeing this question, my mind stray toward physics. in physics we were though that work done = K.E = P.E. where:
work done = force * distance
from this equation:
force =mass(kg)*acceleration(m/s^2) force = newtons (N)
where: mass is measured in kg acceleration is measured in m/s^2
work done = newtons(N)* distance(m) work done = N/m or Joules (J)
the definition of Joules states the work done to move one newton a distance of one meter.
this implies that if work done is equal to K.E. therefor the units for K.E must be in Joules. therefore given:
K.E = 1/2 m v^2 where:
m= mass in kilograms(kg) v= velocity (m/s)
therefore:
K.E = 1/2 (kg) (m/s)^2
K.E = 1/2 (kg) (m/s^2)(m)
if u recall a previous equation, u will of noticed that: (kg) * (m/s^2)= force = newtons (N)
therefore: K.E = 1/2 (kg) (m/s^2)(m) is actually
K.E = 1/2 (N)(m)
and u will may notice that N*m = Joules(J)
therefore from this it was proven that:
K.E = Joules = work done
and this may have shown why K.E is directly proportional to velocity to a power * mass. it is so because it must supplement the law that work done = energy change (K.E)
in an other case, without using the laws of physics, if: K.E = 1/2 m v^2 this means that if drawn on a graph,'K.E' vs 'v' where 'm' was constant, the values will produce a curve with a gradient of:
d(K.E)/d(v)= mv which is also known as momentum in physics. therefore the gradient of the curve a any point on this curve could be found by using:
grad function = mv where: m= mass(kg) v= velocity(m/s)
but this same curve can be converted into a straight line by introducing logs and this straight line would satisfy K.E. is proportional to the velocity to a power by the mass. so when logs is introduced the equation changes to:
log(K.E)= log(0.5mv^2) log(K.E) = log(v^2) + log(0.5m) log(K.E) = 2log(v) + log(0.5m) and this equation is of the form:
Y=mX + c
where: Y = log(K.E) m = 2 X = log v c= log(0.5m)
when this is plotted on a graph, a straight line is formed, and thereby satifying that K.E. is proportional to the velocity to a power by the mass.
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44 comments:
hmmmmmm. miss u taxin men brain wih this question any how i go try
so K.E. proportional to m v^x
let us say the power is x
K.E. = kmv^x
log K.E. = log kmv^x
log K.E. = log k + log mv^x
log K.E. = log k + log m + log v^x
whey ah stick i probably goin wrong
someone
HELP
moose
why it is u bring logs into the equation? u confusing me.which part the 1/2 gone?
miss i think there's an error in the post. K.E is proportional to half the velocity to a power by the mass.
ok moose u kinda on the rite track
wel i would try to help u out...wel actually help myself
ok Ke is porpotional to mv^x then if u introduce an = sign then u need to introduce a constant k... im with u up to here... now my question......how did u get logs??? u just use logs jus so ???
i think i would have used graphs an different set of values in small increments..e.g 0.0000001, 0.000001 ,0.00001 ect..
CAPTAIN VEGETABLE!!! HOW COME U EH POST ANYTING ON THIS ONE AS YET. LIKE YUH SUFFER FROM A NERVOUS BREAKDOWN OH WAT????? HAVE SOME MEAT!!! AHHHHH!!! HAAAA!! HA!! HA HA HA!!!!
Anyway back to more important stuff!!! Ha Ha Ha Ha!!! Oh Gosh!!!
Enough!!! Enough!!
From this question I cud gather that K.E directly prop to
mv^p so K.E = kmv^p. Im kinda lost. Any ideas CAPTAIN VEGTABLE!! I KINDA NEED U!!! ARE U DERE???? Moose why u take logs????
i'm a bit confused too why are you bringing logs into this please explain thank you
now in a scientific experiment there will be the use of a graph for this discovery.
and it is only through the means of a graph showing results of experiments carried will this question be solved
oh and spartan if u read the question u would see that the post said that ke is proprtional to THE VELOCITY TO A POWER MULTIPLY BY MASS THERE WAS NO MENTION OF 1/2 OR THE POWER OF 2
well, saying that K.E. is proportional to velocity to a power by a mass would be represented as:
K.E. = k(v^x)m
but a value must be present to solve for k and then you differentiate, or you don't differentiate, or i probably on the wrong topic, WELL??
spartan, where you seeing 1/2? no variables or values were given?
Moose, i really don't see any relationship between this equation and logs. however if i'm wrong i stand to be corrected!!!
dmcxlKE: mv^2/2...KE: mv^2/2.........KE: mv^2/2....KE: mv^2/2...HOW HOW HOW!!! IM CLUELESS!! HELP!!!!!!
EVERYBODY TALKING ABOUT LOGS!!! HOW LOGS CUM IN DIS????
i really don't know how to do it in logs but i can derive it in a next way(using motion equations)but we didn't do that in class so ...... can someone help us all with this question?
i have no idea how to derive this... but i can sure use it to find angular velocity in a solid wheel... rasif never showed us how it was derived i think
waaaaay miss!! u tryin to bun out we brains! :)
anyway, this formula is used in physics a lot, so instead of using logs and so on, can't we just use other formulas depicting force or velocity and just substitute parts from them into one formula?
what i mean is... instead of using logs and graphs and all those things, if we use certain elements of simple formulas such as 'f=ma' where 'a = change in v / t' and so on, won't we be able to get back the same formula?
plez tell me if i'm totally off. :)
moose y u bring in logs for it looks more intimidating now. i would like som help in clearin up this plz. im not 2 good in physics and bringing it into maths is jus making more trouble for me so can i get som help
first of all what is kinetic energy?
kinetic energy is used to describe the amount of energy a moving energy that a body contains.
so what about an object that is subjected to free fall. since the acceleration due to gravity is 9.8m/s[2], how can the kinetic energy of the body be calculated.
anyhow about the formula;
K.E is proportional to v^x multplied by the mass.
first of all multiplying the momentum by the velocity gives the formula.
another way is the integral of v multiplied by the the mass.
however i don't know how to explain it any further.
@ spartan 117
i think its possible.we may be able to change aroung de formulars...but then if we do that what about de odda people who never did physics ? they wouldnt understand any of it...we would have to go into alot of xplaination..
@ yoshi
what is integral? what does it means? i confuse day papi..so xplain dat concept
wait ent the purpose of logs was to help calculate the product or multiplication of 2 large numbers
I dont understand how come alyuh using logs...and no one is answering the question that everyone is asking...why start off with logs?
fredbugger i think they using logs here coz ent scientist does try all kinda methods to solve problems ? well rite dem wanna be scientists so let them be!!! lol
so if they try all kinda methods then they can use differentation to try and solve this then? or the differentation of logs perhaps ?
wel may be.. i for one aint no scientist so i dunno nah..i jus rember dat miss say they does use small changes in the values when plotting graphs...like 0.00001 , 0.0001 and so on...
if we use certain formulas such as 'f=ma' where 'a = change in v / t' and so on, won't we be able to get back the same formula?
f=ma where a = change in v / t
K.E is proportional to v^x multplied by the mass.
first of all multiplying the momentum by the velocity gives the formula.
another way is the integral of v multiplied by the the mass.
This is my version for this problem. If I am a scientist. I would most likely have to prove this using a graph with a curve.In order to use the graph to get information I would have to use logs.
take logs on both sides
log K.E= -2log m + 2log v
log K.E= log m^(-2) + log v^(2)
log K.E= log m^(-2)x v^(2)
log K.E= log m/2 x v^(2)
log K.E= log 1/2 x m x v^(2)
drop logs
K.E = 1/2 m v^(2)
this question is a bit tricky so i'll try my best and see if i can do this,
so if K.E. is proportional to the velocity to a power
we have:
K.E.= k v^x
where k is a constant of proportionality, v is velocity and x is its unknown power. since this is then multiplied by a mass then the formula should be
K.E. = kmv^x
I know it a bit different from what every one else is saying with all this logs and integral but i think this is the basic formula
to n2o:
you mind explaining how u get the first part of your equation plez.
(-2log m + 2log v)
well let me see--
But moose why you take logs for???
If that is how it really started I lost that is going to defeat the purpoes of what I ahave in mind.
K.e is movement due to the movement of the mass times 1/2 the velocity^2 .
If i was a scientist, i would say well k.e depends on movement and movement depends on speed or velocity. K.E. is directly proportional to v^2 this meaning as velocity increases the energy would also increase, assuming that mass is constant.OK! but linear momentum is also proportional on velocity. so i would think that the power plays a very important role but what is it. if u differentiate K.E. u would get linear momentum, which is mv so we can se there is a relationship. ask urself what is linear momentum? i think momentum is the power residing in a moving object. The thing is mv presents itself graphically as a straight line but 1/2mv^2 is a curve...i will get bak to u on this miss, i hav to go to ur class. i need to brain storm some more
Ok...i tink papa vigil has opened up a door here. i dont know if im correct on this one but he talked about a straight line. Remember momentum is equal to mass by velocity. remember mass is constant. so if we plot a graph of momentum against velocity we wud get a straight line. I tink the gradiant of this line suppose to be 2 and the y intercept suppose to be 1/2. sooooo..um we can now put this in the form ax^n..ENT??? wat u all tink?????
RATMAN i used logs to get rid ofthe power in the equation
Ke= kmv^x
where k is the constant of proportionality, m is the mass and x is the unknown power.
taking out the constant we get
KE= mv^x
i think that the relationship can be calulated experimentally.
then write the equations keeping each variable constant then manipulating it to get the factor by which they should be multiplied.
thought about it some more miss, and where i stuck b4 was because i was thinking why would they square v in the first place...why it could not be 1/3mv^3 or 1/4mv^4...but then i realised that when u square v, ur units will be (m/s)^2 which is (m^2/s^2), which can be seen to be m(ms^-2)...do u see that u units would infer that there is a distance by a constant acceleration...then i thought P.E=K.E., so the the P.E a body has, at a height is equal to the K.E it has as it travels throuh that distance...so i would presume as the scientist, that v^2 is the constant acceleration by that distance as mentioned b4..the coefficient of 1/2 is used to get mv when 1/2mv^2 is differentiated
this qusetion has notting to do with logs . first gust dress up the given equation to the required equation. Be very carefull of where u apply your rules.
But before u do any calculations, Remember u are a scientist, and u have to work on the concepts first, so u will plot a graph and fing the constant, or make any diductions that is usefull.
Now u may solve the equation:
K.E. = kmv^x Thin may tie u up because so you put:
K.E. = kmv
K.E. = km(v^2)^1/2
u then square root the V and after u square V to get rid of the square root.And u will end up with the answer bellow
K.E. = kmv^2
Then u take logs now:
then:logK.E. = logk logm logv^x
logK.E. = logk^k logm logv^x
logK.E. = 1 logm logv^x
take of the logs now:
logK.E. = 1 log(mv^x)
K.E. = 1m^1/2 * v^2
K.E. = 1/2mv^2.
miss, i like this question! because it makes u think about concepts and not just practical formulatiion...captain vegetable why did u not post anything 4 this question, is this question too high of a caliber than ur mind's capacity can handle
according to papa vigil and to add what he say but notin the rouhg sense captain veg if u could answer this question please do
if you have the knowledge please share it for curious blogbakes like me it will be much appreaciated
oh and captain vegetable i hope yuh belly not thin because it seems no one on the blog likes u but
for me u does make me real think so i guess u make sense
copy cat could u explain this part again please i dont understand:
'
u then square root the V and after u square V to get rid of the square root.And u will end up with the answer bellow
K.E. = kmv^2
Then u take logs now:
then:logK.E. = logk logm logv^x
logK.E. = logk^k logm logv^x
logK.E. = 1 logm logv^x
take of the logs now:
logK.E. = 1 log(mv^x)
K.E. = 1m^1/2 * v^2
K.E. = 1/2mv^2.'
In continuation to my previous comment, i would establish that Force= mass x acc. and energy or work = force x dist...so from the equation k.e= 1/2mv^2, u actually hav a mass(m), an acc.(ms^-2), a dist.(in meters).
why do u need CAPTAIN VEGETABLE when the master is here. seeing this question, my mind stray toward physics. in physics we were though that work done = K.E = P.E. where:
work done = force * distance
from this equation:
force =mass(kg)*acceleration(m/s^2)
force = newtons (N)
where:
mass is measured in kg
acceleration is measured in m/s^2
work done = newtons(N)* distance(m)
work done = N/m or Joules (J)
the definition of Joules states the work done to move one newton a distance of one meter.
this implies that if work done is equal to K.E. therefor the units for K.E must be in Joules.
therefore given:
K.E = 1/2 m v^2
where:
m= mass in kilograms(kg)
v= velocity (m/s)
therefore:
K.E = 1/2 (kg) (m/s)^2
K.E = 1/2 (kg) (m/s^2)(m)
if u recall a previous equation, u will of noticed that:
(kg) * (m/s^2)= force = newtons (N)
therefore:
K.E = 1/2 (kg) (m/s^2)(m)
is actually
K.E = 1/2 (N)(m)
and u will may notice that
N*m = Joules(J)
therefore from this it was proven that:
K.E = Joules = work done
and this may have shown why K.E is directly proportional to velocity to a power * mass. it is so because it must supplement the law that
work done = energy change (K.E)
in an other case, without using the laws of physics,
if:
K.E = 1/2 m v^2
this means that if drawn on a graph,'K.E' vs 'v' where 'm' was constant, the values will produce a curve with a gradient of:
d(K.E)/d(v)= mv
which is also known as momentum in physics.
therefore the gradient of the curve a any point on this curve could be found by using:
grad function = mv
where:
m= mass(kg)
v= velocity(m/s)
but this same curve can be converted into a straight line by introducing logs and this straight line would satisfy K.E. is proportional to the velocity to a power by the mass.
so when logs is introduced the equation changes to:
log(K.E)= log(0.5mv^2)
log(K.E) = log(v^2) + log(0.5m)
log(K.E) = 2log(v) + log(0.5m)
and this equation is of the form:
Y=mX + c
where:
Y = log(K.E)
m = 2
X = log v
c= log(0.5m)
when this is plotted on a graph, a straight line is formed, and thereby satifying that K.E. is proportional to the velocity to a power by the mass.
I DARE CAPTAIN VEGETABLE TO GET BETTER THAN THAT..... AS ONE OF MY FRIEND SAY HAA HA YUH BAKE/
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