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Work, Kinetic energy, and power question any help?

by Puk3s
Tags: energy, kinetic, power, work
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Puk3s
#1
Oct20-10, 11:33 PM
P: 3
1. The problem statement, all variables and given/known data

To complete your master's degree in physics, your advisor has you design a small, linear accelerator capable of emitting protons, each with a kinetic energy of 11.3 keV. (The mass of a single proton is 1.67*10^(-27) kg.) In addition, 1.00*10^9 protons per second must reach the target at the end of the 1.80-m-long accelerator.

(a) What the average power must be delivered to the stream of protons?
_____μW

(b) What force (assumed constant) must be applied to each proton?
______ N

(c) What speed does each proton attain just before it strikes the target, assuming the protons start from rest?
_______ m/s

2. Relevant equations

Work - kinetic energy theorm
Work formula


3. The attempt at a solution

Ok so I said KE= Work total or KE=Fd. I have all of those to get the force for question 2, then plug it into my power equation P=Fv for number 1 then W=1/2mv^2 for number 3... but I'm all mixed up on the units... so any help on that would be great. And I'm not so sure my way is right either...
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quenderin
#2
Oct25-10, 05:19 PM
P: 17
Assuming the protons begin at rest, every second you have to deliver 1E9 protons at 11.3keV each. How much energy do you need to deliver per second, i.e. how much power?

If you had a constant force acting on the proton, then you would have a constant acceleration. You know what the final velocity of the proton should be (since you have the final kinetic energy). You also have the distance over this acceleration needs to be done. HOw can you find the required acceleration?

Remember the conversion between eV and J: 1.6 x 10-19 J = 1 eV.
chikenchess
#3
Oct25-10, 11:57 PM
P: 2
Please help out with this question i'm quite confused.This question was given to me by my lecturer but i got a different annswer from what the class got.i need some explanations please.
Question is, A force of 2N gave an object a velocity of 0.028m/s.Calculate the power of the system.
MY APPROACH
Since the object was in motion the energy is Kinetic Energy (KE).Now Power=Work(Energy in this case)/time ie.P=(w/t) but w=K.E hence P=(K.E)/t
Also i know that K.E=0.5mv^2 hence P=0.5mv*(v/t) where (v/t)=acceleration
implying P=0.5mav.
But ma=F therefore P=0.5*2*0.028 P=0.028W
but my class used P=F*(s/t) implying P=Fv hence they got, P=2*0.028 P=0.056
my question once again is why should these answers differ if Power can be expressed in terms of these two methods of derivation?

quenderin
#4
Oct26-10, 12:05 AM
P: 17
Work, Kinetic energy, and power question any help?

I assume the question is to find the Power of the system at the instant when a force of 2N is acting on a particle moving at 0.028 m/s. The power is work done divided by time, NOT kinetic energy of the particle divided by time. The kinetic energy of the particle is equal to the work that HAS BEEN done on it to bring it up to the speed it has at the moment. So what you want is actually the force at this instant times a small change in distance divided by the corresponding small change in time, which is P = Fv.
Mindscrape
#5
Oct26-10, 12:12 AM
P: 1,874
The problem with your approach chiken, is that you didn't take a derivative. P is dE/dt (if we invoke work energy theorem as we can in this case), so that 1/2 goes away when you take the derivative. Or, you could alternatively (for a constant force) simply absorb the derivative into the F*ds, and get velocity i.e. P=Fv.

Even with all that being said, you have to get the average power right in the first step, and the force right in the second step.
chikenchess
#6
Nov21-10, 03:09 PM
P: 2
Okay quendrin and mindscrape thanks guys i got it and i really appreciate it.thanks one more


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