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
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
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.
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