Posted by **carlton** on Monday, October 17, 2011 at 10:47am.

An electron of mass 9.11 10-31 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 = (8-3)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 * 10-22 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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