Question

An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 1.25 cmcm away. It reaches the grid with a speed of 3.3×106 m/sm/s. The accelerating force is constant. If the accelerating force is constant, compute (a) the acceleration; (b) the time to reach the grid; (c) the net force, in newtons.

Answer 1

Step-by-step explanation:

We will use the equations of constant acceleration to find out
`a_(x)` and time t.

As we know that the initial speed is zero. So

(a)

`v_(0x) = 0`

`x - x_(o) = 1.25`×
`10^(-2)`m

`v_(x) = 3.3`×
`10^(6)`m/s

`v^(2) _(x) = v^(2) _{x_(o) } + 2a_(x) (x - x_(o) )`

`a_(x) = (v^(2) _(x) - v^(2) _(ox) )/(2(x - x_(o)) )`

=
`((3.3 * 10^(6))^(2) - 0 )/(2(1.25 * 10^(-2)) )`

= 4.356×
`10^(14)` m/s²

(b)

`v_(x) = v_(ox) + a_(x)t`

`t = v_(x) - vo_(x)/a_(x)`

`t = (3.00 * 10^(6) )/(4.356*10^(14) )` = 6.8870×
`10^(-9)`s

(c)

Σ
`F_(x) = ma_(x)`

= (9.11×
`10^(-31)`)(4.356×
`10^(14)`m/s²)

= 3.968×
`10^(-16)` N