An electron is a subatomic particle (m = 9.11 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +5.16 105 m/s to a final velocity of +2.23 106 m/s while traveling a distance of 0.037 m. The electron's acceleration is due to two electric forces parallel to the x axis: 1 = +6.89 10-17 N, and 2, which points in the -x direction. Find the magnitudes of the net force acting on the electron and the electric force 2.

To find the magnitudes of the net force acting on the electron and the electric force 2, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

First, let's find the acceleration of the electron. We can use the equation of motion:

(vf^2 - vi^2) = 2as

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and s is the distance traveled. Plugging in the given values:

(2.23 * 10^6 m/s)^2 - (5.16 * 10^5 m/s)^2 = 2a * 0.037 m

Simplifying and solving for 'a':

a = [(2.23 * 10^6 m/s)^2 - (5.16 * 10^5 m/s)^2] / (2 * 0.037 m)

a = 1.987 * 10^13 m/s^2

Now, let's find the net force acting on the electron. Using Newton's second law:

Fnet = ma

Plugging in the mass of the electron (m = 9.11 * 10^-31 kg) and the acceleration we just found:

Fnet = (9.11 * 10^-31 kg)(1.987 * 10^13 m/s^2)

Fnet = 1.812 * 10^-17 N

Finally, let's find the magnitude of electric force 2. We know that the net force is the vector sum of electric forces 1 and 2, so:

Fnet = F1 + F2

F2 = Fnet - F1

Plugging in the values:

F2 = (1.812 * 10^-17 N) - (6.89 * 10^-17 N)

F2 = -5.078 * 10^-17 N

The magnitude of the net force acting on the electron is 1.812 * 10^-17 N and the magnitude of electric force 2 is 5.078 * 10^-17 N.