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 +5.21 105 m/s to a final velocity of +2.29 106 m/s while traveling a distance of 0.034 m. The electron's acceleration is due to two electric forces parallel to the x axis: 1 = +7.03 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.

Question

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 +5.21 105 m/s to a final velocity of +2.29 106 m/s while traveling a distance of 0.034 m. The electron's acceleration is due to two electric forces parallel to the x axis: 1 = +7.03 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.

Answer 1

KE change = 2 a X

X = 0.034 m
Solve for a

Fnet = m*a

Fnet = F1 - F2

You know F1; solve for F2

Answer 2

To find the magnitudes of the net force acting on the electron and the electric force 2, we first need to calculate the electron's acceleration.

The initial velocity (u) is +5.21 × 10^5 m/s, the final velocity (v) is +2.29 × 10^6 m/s, and the distance (s) traveled is 0.034 m.

We can use the following kinematic equation to find the acceleration (a):

v^2 = u^2 + 2as

Rearranging the equation, we have:

a = (v^2 - u^2) / (2s)

Substituting the given values:

a = ((2.29 × 10^6)^2 - (5.21 × 10^5)^2) / (2 × 0.034)

Now we can calculate the acceleration:

a = (5.2441 × 10^12 - 2.70841 × 10^11) / 0.068
= 4.97369 × 10^13 / 0.068
≈ 7.313 × 10^14 m/s^2

After obtaining the acceleration, we can find the net force acting on the electron by using Newton's second law:

F_net = m * a

Given that the mass (m) of the electron is 9.11 × 10^-31 kg, we can calculate the net force:

F_net = (9.11 × 10^-31 kg) * (7.313 × 10^14 m/s^2)
= 6.66443 × 10^-16 N

Therefore, the magnitude of the net force acting on the electron is approximately 6.66443 × 10^-16 N.

Next, we can find the magnitude of the electric force 2. The net force acting on the electron is the sum of the two electric forces:

F_net = F_1 + F_2

Given that the magnitude of electric force 1 (F_1) is +7.03 × 10^-17 N, and it acts in the +x direction, we have:

F_net = +7.03 × 10^-17 N + F_2

So, we can rearrange the equation to find the magnitude of electric force 2 (F_2):

F_2 = F_net - 7.03 × 10^-17 N

Substituting the calculated value for F_net:

F_2 = 6.66443 × 10^-16 N - 7.03 × 10^-17 N
= 5.96443 × 10^-16 N

Therefore, the magnitude of the electric force 2 is approximately 5.96443 × 10^-16 N.