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An electron of mass 9.11 x 10-31 kg has an initial speed of 3.40 x 105 m/s. It travels in a straight line, and its speed increases to 6.60 x 105 m/s in a distance of 5.80 cm. Assume its acceleration is constant.
(a) Determine the magnitude of the force exerted on the electron.(N)
(b) Compare this force (F) with the weight of the electron (Fg), which we ignored.
F/Fg= ?

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cjrodriguez89

Answer:

a) 2.5*10⁻¹⁸ N b) 2.8*10¹¹ times larger.

Explanation:

a)

  • Assuming it's moving in a straight line, and that the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of a:

        [tex]v_{f}^{2} - v_{0} ^{2} = 2*a*x[/tex]

  • Replacing  by the givens, we can solve for a, as follows:

       [tex]a = \frac{(v_{f}^{2}-v_{0}^{2} )}{2*x} = \frac{(6.6e5m/s) ^{2}-(3.4e5 m/s)^{2} )}{2*0.058m}= 2.8e12 m/s2[/tex]

  • According to Newton's 2nd law, the magnitude of the force exerted on the electron (which cause it to accelerate) is defined by the following expression:

       [tex]F = m*a = 9.11e-31 kg * 2.8e12 m/s2 = 2.5e-18 N[/tex]

b)

  • The weight of the electron can be calculated as follows:

        [tex]F_{g} = m*g = 9.11e-31 kg * 9.8 m/s2 = 8.9e-30 N[/tex]

  • Dividing F by Fg, we get:

        [tex]\frac{F}{F_{g}} = \frac{2.5e-18N}{8.9e-30N} = 2.8e11[/tex]

  • The force (F) is 2.8*10¹¹ times larger than Fg.

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