Posted by carlton on Monday, October 17, 2011 at 10:47am.
An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to 8.00 105 m/s in a distance of 10.0 cm.
(a) Assuming its acceleration is constant, determine the force exerted on the electron.
N (in the direction of motion)
(b) What is the ratio of this force to the weight of the electron, which we neglected?

physics  Damon, Monday, October 17, 2011 at 11:07am
m = 9.11 * 10^31
vi = 3 * 10^5
vf = 8 * 10^5
vf = vi + a t
d = Vi t + .5 a t^2
10^1 = 3*10^5 t + .5 a t^2
or
1 = 3*10^6 t + 5 a t^2
but
t = (83)10^5 /a
so
1 = 3*10^6 (5*10^5/a ) +5 (25*10^10/a)
a = 15*10^11 + 12.5*10^11
a = 27.5 * 10^11 kg
F = ma = 9*10^31*27.5*10^11
= 247.5 * 10^20
= 2.475 * 1022 N
weight of electron = 9.81*9.11*10^31
so do ratio

physics  drwls, Monday, October 17, 2011 at 11:10am
(a) Force * distance = kinetic energy increase
Solve for the force.
(b) Divide first answer by m*g, where m is the electron mass.
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