Finding Velocity in Work Problem

[petern]

[SOLVED] Finding Velocity in Work Problem

I've already solved A and B and I used W = Fxcos(theta) but that has nothing to do with velocity. Can someone please help me on C and D.

 

[Doc Al]

How does work relate to kinetic energy?

 

[rock.freak667]

The work done in by the force from x=0 to x=3 is converted into kinetic energy...
i.e. W=\frac{mv^2}{2}find v

 

[petern]

Solved, thanks!

 

[jhodzzz]

i tried solving the problem considering the formula given but i didn't get the correct answer. could someone give me the detailed solution on solving C and D?? please?? it would really be of great help to me as for my assignment. I'm having the same problem so I'm looking for the solutions on forums... thanks...

 

[jhodzzz]

rock, considering the fact that you're the one who gave the formula, could you help me please?? i sent a message on your account too... please..

 

[rock.freak667]

i tried solving the problem considering the formula given but i didn't get the correct answer. could someone give me the detailed solution on solving C and D?? please?? it would really be of great help to me as for my assignment. I'm having the same problem so I'm looking for the solutions on forums... thanks...

Well if you are having the same problem, then you can post your thoughts on the parts you can't get out. Since you are having a problem with part C, look at my first post. From the graph can you find the work done between x=3 and x=0?

 

[jhodzzz]

please open my attached file here...it is a picture...

actually this is my problem... based on the illustration on my attachment, an electron is attracted on the positive end of the capacitor. when applied by a downward Force F, it moves and reaches at rest at the other end of the capacitor. it then moves back because of repulsion. The unknown now is the initial velocity Vi of the electron when force is applied..

Given:

const e = -1.602 x 10^-19 Coulumb
mass of e= 9.11 x 10^-31
distance d= 1mm or 0.001 m
E=10 N/C (ELECTRIC FIELD)

W=F.d (since downward force only, no angle made)
F=qE, where q is the electron and E is the Electric field

W=qEd
=-1.602 x 10^-19C(10 N/C (0.001m)) cancel C
W= -1.602 x 10^-21 N/m or J


could you help me on this one rock? let me know if there are lacking givens...ayt?

 

[jhodzzz]

i'm sorry for the inconvinience...i'm a newbie here at forums, don't know how to directly show a picture here on my reply yet... don't know how to post mathematical expressions here too.. :( i'll be learning it.. again I'm sorry..please bare with me...

 

[queenofbabes]

I'm not sure I understand your problem without the picture. Typing out equations as you have are fine for simple ones, for complicated ones refer to https://www.physicsforums.com/misc/howtolatex.pdf

 

[Doc Al]

please open my attached file here...it is a picture...

actually this is my problem... based on the illustration on my attachment, an electron is attracted on the positive end of the capacitor. when applied by a downward Force F, it moves and reaches at rest at the other end of the capacitor. it then moves back because of repulsion. The unknown now is the initial velocity Vi of the electron when force is applied..

The problem is not quite clear to me, even with the picture. Please state the problem completely, word-for-word as it was given to you.

 

[jhodzzz]

sorry for the inconvinience... actually this project was given to us verbally... i only saw the illustration while our teacher keeps explaining about the problem...

ok, i'll try to explain it again...

an electron is attracted at the positive end of the capacitor (simply because it is negatively charged). if the electron is applied by a downward force, the electron will reach the negative side of the capacitor. question is what was the initial velocity Vi due to force applied, in order for it to reach the negative side of the capacitor.

so the final velocity must be 0 because the electron will be at rest momentarily on the negative side of the capacitor.

Given:

const e = -1.602 x 10^-19 Coulumb
mass of e= 9.11 x 10^-31
distance d= 1mm or 0.001 m , distance between both ends of the capacitor


i hope you understand the problem already.. if i lack some given, just tell me specifically what are those.. ayt??

 

[Doc Al]

ok, i'll try to explain it again...

an electron is attracted at the positive end of the capacitor (simply because it is negatively charged). if the electron is applied by a downward force, the electron will reach the negative side of the capacitor. question is what was the initial velocity Vi due to force applied, in order for it to reach the negative side of the capacitor.

so the final velocity must be 0 because the electron will be at rest momentarily on the negative side of the capacitor.

Allow me to rephrase the problem: An electron starts out at the positive side of the capacitor with some initial speed V, such that it barely makes it to the negative side before coming to rest and then going back. Find that initial speed.

Is that it?

 

[jhodzzz]

yeah2x..that's it... am i missing some given for you to solve the problem??
how am i going to solve it??

 

[Doc Al]

Hint: Use the Work-Energy theorem to relate the work done by the field to the change in KE of the electron.

(Or you can use conservation of energy, which is another way of expressing the same thing.)

 

[jhodzzz]

ahm, I'm really weak at formulas and theorems...could u give me the formulas that i need for the problem? please...

 

[jhodzzz]

i'm passing this one by tomorrow already and i don't have the luxury of time to research those theorems here and now...so I'm asking a favor...i'll try to research on this one afterwards...promise...it's just that i have to solve this immediately.. please...

 

[Doc Al]

http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr"