To solve this problem, we need to consider the forces acting on the patient during the triceps dip exercise.
The total force exerted by the arms is 155 N on each arm, so the total upward force applied by the arms is 310 N. This force acts over a distance of 25.0 cm, which is equivalent to 0.25 meters.
The work done by the arms is equal to the force multiplied by the distance:
Work = Force x Distance
Work = 310 N x 0.25 m
Work = 77.5 J
This work done by the arms results in kinetic energy for the patient. The total kinetic energy of the patient is given by the formula:
KE = 0.5 * mass * velocity^2
Since the patient's initial velocity is zero, the kinetic energy comes only from the work done by the arms:
KE = Work
KE = 77.5 J
Substitute the total force acting on the patient (700 N) exclusive of the arms into the kinetic energy formula:
KE = 0.5 * (700 N) * velocity^2
Set the two expressions for kinetic energy equal to each other:
0.5 * (700 N) * velocity^2 = 77.5 J
Solve for velocity:
velocity^2 = (77.5 J) / (0.5 * 700 N)
velocity^2 = 0.22 m/s
Therefore, the patient's velocity at this point is approximately 0.47 m/s.
A patient doing the triceps dip exercise weighs 700 N exclusive of the arms. During the first 25.0 cm of the lift, each arm exerts an upward force of 155 N. If the upward movements starts from rest, what is the patient's velocity at this point? (Precaution: If a patient is prone to anterior subluxation or dislocation, this activity should not be done.)
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