Question

An electron of mass 9.11*10-31kg has an initial speed of 3.00*105m/s. It travels in a straight line and its speed increases to 7.00*105m/s in a distance of 5.00cm. Assuming its acceleration is constant, (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

Answer 1

Answer:

a)  F = 36.4 10⁻¹⁹ N   b)  F /W = 4.08 10¹¹

Explanation:

a) To find the force we will use Newton's second law and to find the acceleration we will use the kinematics equations

    $v_{f}$² = v₀² + 2 a x

    a = ($v_{f}$²-v₀²) / 2x

   a = ((7 10⁵)²2 - (3 10⁵)²) / 2 0.05

   a = 4.00 10¹² m /s²

With Newton's second law we calculate the force

   F = m a

   F = 9.11 10⁻³¹ 4 10¹²

   F = 36.4 10⁻¹⁹ N

b) calculate the  weight of the electron

   W = mg

   W = 9.11 10⁻³¹ 9.8

   W = 89.3 10⁻³¹ N

The comparison is made by dividing the two magnitudes

   F / W = 36.4 10⁻¹⁹ / 89.3 10⁻³¹

   F /W = 4.08 10¹¹