An electron in a television tube is accelerated uniformly from rest to a speed of 8.4/times 10^7~/text{m/s}8.4*10 7 m/s over a distance of 2.5 cm. What is the instantaneous power delivered to the electron at the instant that its displacement is 1.0 cm?

P=3.42*10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that:

vf^2=vi^2+2*a (xf-xi)

Where:

• vf, vi, are the the final and the initial velocity of the electron

• a is the acceleration of the electron

• xf, xi are the final and the initial position of the electron.

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.

Givens: vf = 8.4 x 10^7 m/s, vi, = 0 m/s, xf = 0.025 m and xi = 0 m

vf^2 = vi^2+2*a (xf-xi)

vf^2-vi^2=2*a (xf-xi)

2*a (xf-xi) = vf^2-vi^2

a = (vf^2-vi^2) / 2 (xf-xi)

Pluging known information to get:

a = (vf^2-vi^2) / 2 (xf-xi)

= 1.411 * 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m

so,

vf^2 = vi^2+2*a (xf-xi)

vf^2 = 5.312 * 10^7

From the following Eq. we can calculate the time elapsed in this motion.

xf = xi+vi*t+1/2*a*t

xf = xi+vi*t+1/2*a*t

t=√2 (xf-xi) / a

t=3.765*10^-10 s

now we can use the power P Eq.

P=W/Δt = > ΔK/Δt

Where: the work done W change the kinetic energy K of the electron,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42*10^-6 J/s