An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed increases to 6.60

Question

An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed increases to 6.60 x10^5 m/s in a distance of 5.40 cm. Assume its acceleration is constant.

Required:
a. Determine the magnitude of the force exerted on the electron.
b. Compare this force (F) with the weight of the electron (Fg), which we ignored.

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Thu Thủy 5 years 2021-08-22T23:18:21+00:00 1 Answers 37 views 0

Answers ( )

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    2021-08-22T23:19:57+00:00

    Answer:

    a.     F = 2.32*10^-18 N

    b.     The force F is 2.59*10^11 times the weight of the electron

    Explanation:

    a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

    v^2=v_o^2+2ax         (1)

    v: final speed of the electron = 6.60*10^5 m/s

    vo: initial speed of the electron = 4.00*10^5 m/s

    a: acceleration of the electron = ?

    x: distance traveled by the electron = 5.40cm = 0.054m

    you solve the equation (2) for a and replace the values of the parameters:

    a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}

    Next, you use the second Newton law to calculate the force:

    F=ma

    m: mass of the electron = 9.11*10^-31kg

    F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N

    The magnitude of the force exerted on the electron is 2.32*10^-18 N

    b. The weight of the electron is given by:

    F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N

    The quotient between the weight of the electron and the force F is:

    \frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}

    The force F is 2.59*10^11 times the weight of the electron

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