Question

An electron of mass 9.11 x 10⁻³¹ kg has an initial speed of 3.00 x 10⁵ m/s. It travels in a straight line, and its speed increases to 9.00 x 10⁵ m/s in a distance of 4.00 cm. 1). Assuming its acceleration is constant, determine the force exerted on the electron. N (in the direction of motion)

Answer 1

Answer: a. The magnitude of the force exerted on the electron F = 3.64*10^-18N

b. The force F is 4.08*10^11 times the weight of the electron

Explanation: a. The magnitude of the force exerted on the electron F = 3.64*10^-18N

b. The force F is 4.08*10^11 times the weight of the electron

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: final speed of the electron = 7.00*10^5 m/s

vo: initial speed of the electron = 3.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron with certain speed= 5.0cm = 0.05m

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

a=4×10^12 m/s²

to calculate the force we now use the second Newton law :

F=ma

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

F=3.64*10^-18N

The magnitude of the force exerted on the electron is 3.64*10^-18N

b. given weight of electron is :

Fg=mg

Fg = 8.92*10^-30N

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

F/Fg=4.08*10^11

The force F is 4.08*10^11 times the weight of the electron