Question

question 3 an electron is a subatomic particle (m = 9.11 x 10^-31 kg) that is subject to electric forces. an electron moving in the +x direction accelerates from an initial velocity of +5.75 x 10^5 m/s to a final velocity of 2.71 x 10^6 m/s while traveling a distance of 0.0601 m. the electrons acceleration is due to two electric forces parallel to the x - axis: vector f1 = 8.92 x 10^-17 n, and vector f2, which points in the -x direction. find the magnitudes of (a) the net force acting on the electron and (b) the electric force vector f2.

Explanation:

Step1: Use kinematic - equation to find acceleration

First, use the kinematic equation $v^{2}=v_{0}^{2}+2ax$. Rearranging for acceleration $a$, we get $a=\frac{v^{2}-v_{0}^{2}}{2x}$. Given $v = 2.71\times10^{6}\text{ m/s}$, $v_{0}=5.75\times10^{5}\text{ m/s}$ and $x = 0.0601\text{ m}$.
\[a=\frac{(2.71\times 10^{6})^{2}-(5.75\times 10^{5})^{2}}{2\times0.0601}\]
\[a=\frac{7.3441\times 10^{12}- 3.30625\times 10^{11}}{0.1202}\]
\[a=\frac{7.013475\times 10^{12}}{0.1202}\approx5.835\times 10^{13}\text{ m/s}^2\]

Step2: Use Newton's second - law to find net force

According to Newton's second - law $F_{net}=ma$. Given $m = 9.11\times10^{-31}\text{ kg}$ and $a\approx5.835\times 10^{13}\text{ m/s}^2$.
\[F_{net}=(9.11\times10^{-31})\times(5.835\times 10^{13})\]
\[F_{net}\approx5.326\times 10^{-17}\text{ N}\]

Step3: Find the second electric force

We know that $F_{net}=F_1 + F_2$. Given $F_1 = 8.92\times10^{-17}\text{ N}$ and $F_{net}\approx5.326\times 10^{-17}\text{ N}$. Since $F_2$ points in the negative $x$ - direction, we have $F_{net}=F_1+F_2$. Rearranging for $F_2$, we get $F_2=F_{net}-F_1$.
\[F_2 = 5.326\times 10^{-17}-8.92\times 10^{-17}=- 3.594\times 10^{-17}\text{ N}\]
The magnitude of $F_2$ is $3.594\times 10^{-17}\text{ N}$

Answer:

(a) The magnitude of the net force is $5.33\times 10^{-17}\text{ N}$ (rounded to three significant figures)
(b) The magnitude of the electric force $F_2$ is $3.59\times 10^{-17}\text{ N}$ (rounded to three significant figures)