Answer:
[tex]W=\Delta KE= KE_f-KE_i=\frac{1}{2} \,m \,v_f^2-\frac{1}{2} \,m \,v_i^2[/tex]
Explanation:
The Work-Kinetic Energy Theorem states that the work done on a particle of mass "m", equals the particle's change in Kinetic Energy (final Kinetic Energy of the particle "[tex]KE_f[/tex]" minus the initial Kinetic energy of the particle "[tex]KE_i[/tex]"), and it is expressed as:
[tex]W=\Delta KE= KE_f-KE_i=\frac{1}{2} \,m \,v_f^2-\frac{1}{2} \,m \,v_i^2[/tex]
where we have used the explicit form of the KE of a particle of mass m and velocity [tex]v[/tex]. Of course, [tex]v_f[/tex] stands for the final velocity of the particle, and [tex]v_i[/tex] for the particle's initial velocity.
