Question

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 electron acceleration is due to two electric - forces parallel to the x - axis: f₁ = 8.92 x 10^(-17) n, and f₂, which points in the -x direction. find the magnitudes of (a) the net force acting on the electron and (b) the electric force f₂.

Explanation:

Step1: Use the kinematic - equation to find acceleration

We use the kinematic equation $v^{2}=v_{0}^{2}+2ax$. Given $v_{0}=5.75\times 10^{5}\ m/s$, $v = 2.71\times 10^{6}\ m/s$ and $x = 0.0601\ m$. Rearranging for $a$ gives $a=\frac{v^{2}-v_{0}^{2}}{2x}$.
\[

$$\begin{align*} a&=\frac{(2.71\times 10^{6})^{2}-(5.75\times 10^{5})^{2}}{2\times0.0601}\\ &=\frac{7.3441\times 10^{12}- 3.30625\times 10^{11}}{0.1202}\\ &=\frac{73.441\times 10^{11}-3.30625\times 10^{11}}{0.1202}\\ &=\frac{70.13475\times 10^{11}}{0.1202}\\ &\approx5.84\times 10^{13}\ m/s^{2} \end{align*}$$

\]

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

According to Newton's second - law $F_{net}=ma$. Given $m = 9.11\times 10^{-31}\ kg$ and $a\approx5.84\times 10^{13}\ m/s^{2}$. So $F_{net}=ma=(9.11\times 10^{-31})\times(5.84\times 10^{13})\approx5.32\times 10^{-17}\ N$.

Step3: Find the force $\vec{F}_{2}$

We know that $F_{net}=F_{1}-F_{2}$ (taking the $+x$ direction as positive). Given $F_{1}=8.92\times 10^{-17}\ N$ and $F_{net}\approx5.32\times 10^{-17}\ N$. Rearranging for $F_{2}$ gives $F_{2}=F_{1}-F_{net}$.
\[

$$\begin{align*} F_{2}&=8.92\times 10^{-17}-5.32\times 10^{-17}\\ &=3.6\times 10^{-17}\ N \end{align*}$$

\]

Answer:

(a) $5.32\times 10^{-17}\ N$
(b) $3.6\times 10^{-17}\ N$