# Review. An electron of mass 9.11 \times 10−31 kg has an initial speed of 3

Review. An electron of mass $9.11×10-31$ kg has an initial speed of 3.00 $3×105\frac{m}{s}$. It travels in a straight line, and its speed increases to $7.00×105\frac{m}{s}$ in a distance of 5.00 cm. Assuming its acceleration is constant, (a) determine the magnitude of the force exerted on the electron and (b) compare this force with the weight of the electron, which we ignored.
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Ste 1
Given data:
$m=9.11×{10}^{-31}kg$
${v}_{i}=3\mid ×{10}^{5}\frac{m}{s}$
${v}_{f}=7×{10}^{5}\frac{m}{s}$
$\mathrm{△}x=5cm$

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Melinda McCombs
Step 2
(a)
First: we determine the acceleratoin by using the following equation:
${v}_{\left\{f\right\}}^{2}={v}_{\left\{i\right\}}^{2}+2a\mathrm{△}x$
Solving for (a) $a=\frac{{x}_{\left\{f\right\}}^{2}-{v}_{\left\{i\right\}}^{2}}{2\mathrm{△}x}=\frac{\left(7×{10}^{5}\right)-{\left(3×{10}^{5}\right)}^{2}}{2×0.05}=4×{10}^{12}\frac{m}{{s}^{2}}$
Then: we use Newtons
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Don Sumner

Step 3
(b)
The weight of the electron is determined by:
$Fg=mg=\left(9.11×{10}^{-31}\right)\left(9.8\right)=8.93×{10}^{-30}N$
Therefore, this force is larger that the weight of the electron by $\frac{3.64×{10}^{-18}}{8.93×{10}^{-30}}=4.08×{10}^{11}$ times.
So,
(a) $\sum F=3.64×{10}^{-18}N$
(b)This force is larger that the weight of the electron by $4.08×{10}^{11}$ times.