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.
a=(v²-v₀²)/2•s = {(2.23•10⁶)²-(5.16•10⁵)}/2•0.037 =6.36•10^13 m/s²
F(net) =ma= 9.1•10^-31•6.36•10^13=5.78•10^-17 N.
ma=F1-F2.
F2=F1-ma=6.89•10^-17-5.78•10^-17 = 1.11•10^-17 N.(in –x-direction)
To find the magnitudes of the net force acting on the electron and 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.
1. Calculate the acceleration of the electron:
Using the given initial and final velocities and the displacement, we can use the equation of motion:
final velocity squared = initial velocity squared + 2 * acceleration * displacement
(2.23 * 10^6)^2 = (5.16 * 10^5)^2 + 2 * acceleration * 0.037
Solving this equation will give us the acceleration of the electron.
2. Calculate the net force:
Now that we have the acceleration, we can calculate the net force acting on the electron using Newton's second law:
net force = mass * acceleration
The mass of the electron is given as 9.11 * 10^-31 kg.
3. Calculate the electric force 2:
To find the electric force 2, we need to find the sum of the forces acting on the electron in the x-direction.
Since we have the magnitude and direction of force 1, we can subtract it from the net force to find the magnitude of force 2.
net force = force 1 + force 2
Solving this equation will give us the magnitude of force 2.
By following these steps, you should be able to find the magnitudes of the net force acting on the electron and the electric force 2 in the given scenario.
To find the magnitudes of the net force acting on the electron and the electric force 2, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
Step 1: Find the acceleration of the electron:
The initial velocity of the electron, v1 = +5.16 × 105 m/s.
The final velocity of the electron, v2 = +2.23 × 106 m/s.
The displacement of the electron, Δx = 0.037 m.
We can use the equation v2^2 = v1^2 + 2aΔx, where "a" is the acceleration.
Solving for "a", we have:
a = (v2^2 - v1^2) / (2Δx)
= ((2.23 × 10^6)^2 - (5.16 × 10^5)^2) / (2 × 0.037)
= 6.31 × 10^12 m/s^2
Therefore, the electron's acceleration is 6.31 × 10^12 m/s^2.
Step 2: Find the net force acting on the electron:
Using Newton's second law, F_net = ma, where "m" is the mass of the electron.
The mass of the electron, m = 9.11 × 10^-31 kg.
The net force acting on the electron is:
F_net = m × a
= 9.11 × 10^-31 kg × 6.31 × 10^12 m/s^2
= 5.75 × 10^-18 N
Therefore, the magnitude of the net force acting on the electron is 5.75 × 10^-18 N.
Step 3: Find the magnitude of electric force 2:
The magnitude of electric force 1, F1 = 6.89 × 10^-17 N.
Since the net force acting on the electron is the vector sum of the two electric forces, we can write:
F_net = F1 + F2
Rearranging the equation, we find:
F2 = F_net - F1
= 5.75 × 10^-18 N - 6.89 × 10^-17 N
= -6.31 × 10^-17 N
Therefore, the magnitude of electric force 2 is 6.31 × 10^-17 N.