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thepatient
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So there is this work-energy problem online that I was discussing with a friend. We both concluded with different answers and I 99.99% sure that I am correct, but there is that 0.01% chance that my understanding of physics has been completely wrong over the years. The problem is simple:
A moving electron has kinetic energy K1 . After a net amount of work W has been done on it, the electron is moving one-quarter as fast in the opposite direction. Find W in terms of K1.
I came up with the answer:
W = -15/16 K1
He came up with:
-17/16 W1
My reply to his answer:
Work is defined to be the change in kinetic energy, why would KE, and not the change of it, be equal to work? Explain please.
And also, when traveling in one direction, and when there's a force acting in the opposite direction and slowing you down, you are not acquiring KE but losing KE, negative work is being done. Why do you say KE acquired is 1/16 of work? Explain please because I was pretty sure I was write, but I can be wrong. XD
His reply:
If W1 was, say 100J of KE, then 100J of work must be done (one way or another) - against the motion to bring the object to rest. From rest, the object has to re-gain an amount of KE ... so the total energy needed is going to be more than 100J (W1) ... 15/16 W1 is less than W1.
the additional KE to be gained will only be 1/16 of W1 because the new velocity is now 1/4 of it's original value ... and KE is proportional to v² ...ie ( 1/4 ² = 1/16 )
All the energy requirements (to stop and to move backwards) must be added together irrespective of direction - they are all 'in addition to each other')
Energy is changed as a consequence of doing work... and the amount of energy changed is equal to the amount of work done (or vice versa) - that's the work-energy 'theorem'. Both work and energy are measured in Joules to enable this.
The minus sign isn't needed, I used it to indicate that this work was to be done in the opposite direction to the original motion.
hope this is helpful
My reply to that:
The thing is that initially it is moving at a constant speed. If an object moves at a constant speed then there is a kinetic energy, but there is no work being done; that is the flaw that I see behind the reasoning.
Remember from Newtons second law that the net force is equal to it's mass times acceleration? if this particle is moving at a constant speed then the net forces acting on the object is zero.
Work is also defined as the dot product of the Force and the Position vectors. If initially F = 0 then W = F * d = 0. There is no work done until the object has a force that moves it to the opposite direction.
Also the work-energy theorem states that, with conservation of energy, the initial energy is equal to the final energy. Energy in = energy out; not the sum of the absolute value of all energies. So in that case Ki + W = Kf, no work by friction or nonconservatve force. Using the work energy theorem we would get that W = Kf -Ki, not the sum of the absolute values of both energies.
Also knowing that work is a dot product of F*d vectors, we know work is a scalar product, so it doesn't have a direction, you can't put a negative sign because it's moving in the negative direction.
And he hasn't replied yet.
I was 100% sure, but is my understanding on the work-energy theorem completely wrong? Who is right and who is wrong? Thanks. :]
A moving electron has kinetic energy K1 . After a net amount of work W has been done on it, the electron is moving one-quarter as fast in the opposite direction. Find W in terms of K1.
Ah and also this here is my work:
K1 = 1/2 mv1^2 <--by definition
v2 = -v1/4 <-- given in the question that the final velocity is a fourth of the initial velocity but opposite in direction.
K2 = 1/2 mv2^2 (when I say v1 and v2 I mean first and final velocity) By definition
Substituting v2 = -v1/4 for the formula of final kinetic energy (to leave it in terms of K1)
K2 = 1/2 mv1^2/4^2 = 1/2 mv1^2/16 = 1/16 K1 <-in terms of K1
By definition W = ∆K
W = ∆K = K2 - K1 = 1/16 K1 - K1 = -15/16 K1 is the total work in terms of K.
A moving electron has kinetic energy K1 . After a net amount of work W has been done on it, the electron is moving one-quarter as fast in the opposite direction. Find W in terms of K1.
I came up with the answer:
W = -15/16 K1
He came up with:
-17/16 W1
My reply to his answer:
Work is defined to be the change in kinetic energy, why would KE, and not the change of it, be equal to work? Explain please.
And also, when traveling in one direction, and when there's a force acting in the opposite direction and slowing you down, you are not acquiring KE but losing KE, negative work is being done. Why do you say KE acquired is 1/16 of work? Explain please because I was pretty sure I was write, but I can be wrong. XD
His reply:
If W1 was, say 100J of KE, then 100J of work must be done (one way or another) - against the motion to bring the object to rest. From rest, the object has to re-gain an amount of KE ... so the total energy needed is going to be more than 100J (W1) ... 15/16 W1 is less than W1.
the additional KE to be gained will only be 1/16 of W1 because the new velocity is now 1/4 of it's original value ... and KE is proportional to v² ...ie ( 1/4 ² = 1/16 )
All the energy requirements (to stop and to move backwards) must be added together irrespective of direction - they are all 'in addition to each other')
Energy is changed as a consequence of doing work... and the amount of energy changed is equal to the amount of work done (or vice versa) - that's the work-energy 'theorem'. Both work and energy are measured in Joules to enable this.
The minus sign isn't needed, I used it to indicate that this work was to be done in the opposite direction to the original motion.
hope this is helpful
My reply to that:
The thing is that initially it is moving at a constant speed. If an object moves at a constant speed then there is a kinetic energy, but there is no work being done; that is the flaw that I see behind the reasoning.
Remember from Newtons second law that the net force is equal to it's mass times acceleration? if this particle is moving at a constant speed then the net forces acting on the object is zero.
Work is also defined as the dot product of the Force and the Position vectors. If initially F = 0 then W = F * d = 0. There is no work done until the object has a force that moves it to the opposite direction.
Also the work-energy theorem states that, with conservation of energy, the initial energy is equal to the final energy. Energy in = energy out; not the sum of the absolute value of all energies. So in that case Ki + W = Kf, no work by friction or nonconservatve force. Using the work energy theorem we would get that W = Kf -Ki, not the sum of the absolute values of both energies.
Also knowing that work is a dot product of F*d vectors, we know work is a scalar product, so it doesn't have a direction, you can't put a negative sign because it's moving in the negative direction.
And he hasn't replied yet.
I was 100% sure, but is my understanding on the work-energy theorem completely wrong? Who is right and who is wrong? Thanks. :]
Homework Statement
A moving electron has kinetic energy K1 . After a net amount of work W has been done on it, the electron is moving one-quarter as fast in the opposite direction. Find W in terms of K1.
Homework Equations
The Attempt at a Solution
Ah and also this here is my work:
K1 = 1/2 mv1^2 <--by definition
v2 = -v1/4 <-- given in the question that the final velocity is a fourth of the initial velocity but opposite in direction.
K2 = 1/2 mv2^2 (when I say v1 and v2 I mean first and final velocity) By definition
Substituting v2 = -v1/4 for the formula of final kinetic energy (to leave it in terms of K1)
K2 = 1/2 mv1^2/4^2 = 1/2 mv1^2/16 = 1/16 K1 <-in terms of K1
By definition W = ∆K
W = ∆K = K2 - K1 = 1/16 K1 - K1 = -15/16 K1 is the total work in terms of K.
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Darn it, my sign convention is messed up!