Question

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?

Answer 1

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

Answer 2

(a) Force * distance = kinetic energy increase
Solve for the force.
(b) Divide first answer by m*g, where m is the electron mass.

Answer 3

thank you

Answer 4

thank u