An electron ( mass = 9.11×10 31 kg ) leaves one end of a TV picture tube with ze

Question

An electron ( mass = 9.11×1031 kg ) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.79 cmaway. It reaches the grid with a speed of 2.97×106 m/s . (You can ignore the gravitational force on the electron.)

a) If the accelerating force is constant, compute the acceleration of the electron.

b) If the accelerating force is constant, compute the time it takes the electron to reach the grid.

c)If the accelerating force is constant, compute the net force that is accelerating the electron, in newtons.

Explanation / Answer

Here ,

initial speed , u = 0 m/s

final speed , v = 2.97 *10^6 m/s

d = 0.0179 m

a)

let the acceleration is a

using third equation of motion

v^2 - u^2 = 2 * a * d

(2.97 *10^6)^2 = 2 * 0.0179 * a

a = 2.464 *10^14 m/s^2

the acceleration of electron is 2.464 *10^14 m/s^2

b)

let the time taken is t

v = u +a*t

2.97 *10^6 = 2.464 *10^14 * t

t = 1.205 *10^-8 s

the time taken for the electron is 1.205 *10^-8 s

c)

net force = m*a

net force = 9.11 *10^-31 * 2.464 *10^14

net force = 2.245 *10^-16 N

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