Question

An electron of mass 9.11 10-31 kg has an initial speed of 3.80 105 m/s. It travels in a straight line, and its speed increases to 7.60 105 m/s in a distance of 4.20 cm. Assume its acceleration is constant. (a) Determine the magnitude of the force exerted on the electron. 0.00000000000000000082875492 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. N (b) Compare this force (F) with the weight of the electron (Fg), which we ignored. F Fg =

Answer 1

Answer:

$F=4.7\times10^{-18}N$

$\frac{F}{W}=525702635794$

Explanation:

The acceleration experimented by the electron can be calculated using the formula $v_f^2=v_i^2+2ad$, so we have:

$a=\frac{v_f^2-v_i^2}{2d}$

The magnitude of of the force exerted on the electron can be then obtained using Newton's 2nd Law F=ma.

Putting all together we have:

$F=ma=\frac{m(v_f^2-v_i^2)}{2d}$

And using our values (and that 4.2cm are 0.042m):

$F=\frac{(9.11\times10^{-31}Kg)((7.6\times10^5m/s)^2-(3.8\times10^5m/s)^2)}{2(0.042m)}=4.7\times10^{-18}N$

Then we compare this force with the weight of the electron:

$\frac{F}{W}=\frac{\frac{m(v_f^2-v_i^2)}{2d}}{mg}=\frac{(v_f^2-v_i^2)}{2dg}=\frac{(7.6\times10^5m/s)^2-(3.8\times10^5m/s)^2}{2(0.042m)(9.81m/s^2)}=525702635794$

So this force is 525702635794 times larger than the weight of the electron.