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?

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

(a) Force * distance = kinetic energy increase

Solve for the force.
(b) Divide first answer by m*g, where m is the electron mass.

thank you

thank u

To find the force exerted on the electron, we need to use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the force is unknown, the mass of the electron is given as 9.11 * 10^(-31) kg, and we need to find the acceleration.

We can calculate the acceleration using the kinematic equation:

v^2 = u^2 + 2as

Where,
v = final velocity (8.00 * 10^5 m/s)
u = initial velocity (3.00 * 10^5 m/s)
a = acceleration (unknown)
s = distance traveled (10.0 cm)

First, we convert the distance from cm to meters:
s = 10.0 cm = 0.10 m

Now we substitute the given values into the equation and solve for acceleration:

(8.00 * 10^5 m/s)^2 = (3.00 * 10^5 m/s)^2 + 2a * 0.10 m

64.00 * 10^10 m^2/s^2 = 9.00 * 10^10 m^2/s^2 + 0.20a

Subtracting 9.00 * 10^10 m^2/s^2 from both sides:

55.00 * 10^10 m^2/s^2 = 0.20a

Dividing both sides by 0.20:

a = (55.00 * 10^10 m^2/s^2) / 0.20 = 275 * 10^10 m^2/s^2

Now we have the acceleration. To find the force exerted on the electron, we can use Newton's second law:

F = ma

Substituting the values:

F = (9.11 * 10^(-31) kg) * (275 * 10^10 m^2/s^2)

Calculating:

F = 2.50 * 10^(-20) N

Therefore, the force exerted on the electron is 2.50 * 10^(-20) N in the direction of motion.

Now let's move on to part (b) to find the ratio of this force to the weight of the electron, which we neglected.

The weight of an object is given by the formula:

Weight = m * g

Where,
m = mass of the object (9.11 * 10^(-31) kg)
g = acceleration due to gravity (9.8 m/s^2)

Substituting the values:

Weight = (9.11 * 10^(-31) kg) * (9.8 m/s^2)

Calculating:

Weight = 8.95 * 10^(-30) N

Now we can calculate the ratio of the force to the weight:

Ratio = Force / Weight

Substituting the values:

Ratio = (2.50 * 10^(-20) N) / (8.95 * 10^(-30) N)

Calculating:

Ratio = 2.79 * 10^9

Therefore, the ratio of the force exerted on the electron to its weight is 2.79 * 10^9.