A physical therapy exercise has a person shaking a 5.00kg dumbbell up and down rapidly. If the barbell is moving at 4.5m/s what is the magnitude of the force required to stop it in .333 seconds

Answer 1

impatient much?
Once is enough, thank you.
Repeated postings will not get faster or better responses.

Answer 2

a = change in velocity / change in time
so
a = 4.5 m/s / 0.333 s = 13.5 m/s^2
F = m a = 5 * 13.5 = 67.6 Newtons