Question

An electron in a television tube is accelerated uniformly from rest to a
speed of 8.4·10 7 m/s over a distance of 2.5 cm. What is the power delivered to the
electron at the instant that its displacement is 1.0 cm?
1) 13,6∙10 -6 J; 3) 6,8∙10 -6 J;
2) 2,4∙10 -6 J; 4) 1,6∙10 -6 J

Answer 1

Explanation:

v² = u²+2as,. u= 0

(8.4×10^7)² = 2a×2.5×10^-2

a = 70.56×10^16/5

= 14.112×10^16

= 1.41×10^17 m/s²

given s = 1 cm = 10^-2 m

v² = 2as = 2×1.41×10^17×10^-2

= 2.82× 10^15

change in kinetic energy= (1/2)mv²

= (1/2)×9.109×10^-31×2.82×10^15

= 1.28× 10^-15 joule

since v= u+at , u = 0 ,=> v = at

t = v/a = 5.31×10^7 / 1.41×10^17

= 3.76×10^-10 s

power delivered = 1.28×10^-15 / 3.76×10^-10

= 3.4×10^-6 J/s