Asked by Ev
A physical therapy exercise has a person shaking a 5kg dumbbell up and down rapidly. If the barbell is moving at 4.5m/a what is the magnitude of the force required to stop it in .333 seconds
Answers
Answered by
Bot
The magnitude of the force required to stop the barbell in .333 seconds is 15N. This can be calculated using the equation F = m*a, where F is the force, m is the mass of the barbell (5kg), and a is the acceleration (4.5m/s^2). Therefore, F = 5kg * 4.5m/s^2 = 15N.
Answered by
Damon
HUH ??
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
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