Question

An electrone cathode ray tube of a television set enters a region where it accelerates uniformly from a speed of 28500 m/s to a speed of 4.37 * 10^6 m/s in a distance of 2.01cm. What is it's acceleration?

Answer 1

You need an equation that relates initial velocity, final velocity, displacement and acceleration. Do you know it?

If not you can derive it using energy considerations.

The work done on the particle is equal to its change of energy.
The work,
W = Fd , where F is the force, d the displacement
= mad , as by Newton's 2nd law F = ma

The change in energy is the change in Kinetic energy, KE. Write u for initial velocity, v final velocity, then
change in KE = (1/2)mv^2 - (1/2)mu^2

This must be equal to the work done, hence

mad = (1/2)mv^2 - (1/2)mu^2

which implies

2ad = v^2 - u^2

You can now solve for a.