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

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

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

$a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }$

   = $\frac{(3.3 * 10^{6})^{2} - 0 }{2(1.25 * 10^{-2}) }$

   = 4.356×$10^{14}$ m/s²

(b)

$v_{x} = v_{ox} + a_{x}t$

$t = v_{x} - vo_{x}/a_{x}$

$t = \frac{3.00 * 10^{6} }{4.356*10^{14} }$ = 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