Question

A net force of 1.6×10−15 N acts on an electron over a displacement of 5.0 cm, in the same direction as the net force. (a) What is the change in kinetic energy of the electron? (b) If the electron was initially at rest, what is the speed of the electron? An electron has a mass of 9.1×10−31 kg.

Answer 1

Explanation:

It is given that,

Net force acting on the electron, $F=1.6\times 10^{-15}\ N$

Displacement, d = 5 cm = 0.05 m

(a) Let $\Delta E$ is the change in kinetic energy of the electron. It can be calculated using work energy theorem. Mathematically, it is given by :

$W=\Delta E$

$\Delta E=F\times d$

$\Delta E=1.6\times 10^{-15}\ N\times 0.05\ m$

$\Delta E=8\times 10^{-17}\ J$

(b) Initial speed of the electron, u = 0

Again using the work energy theorem as :

$E=\dfrac{1}{2}m(v^2-u^2)$

$E=\dfrac{1}{2}m(v^2)$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 8\times 10^{-17}}{9.1\times 10^{-31}}}$    

$v=1.32\times 10^7\ m/s$

Hence, this is the required solution.