Asked by Anissa
an 88.kg fireman slides 5.9 meters down a fire pole. He holds the pole, which exerts a 520N steady resistive force on the fireman. At the bottom he slows down to a stop in 0.43m by bending his knees. determine the acceleration while stopping and the time it takes for the fireman to stop after reaching the ground. I need help figuring out how to sole this.
Answers
Answered by
Scott
his energy at the bottom is the change in potential energy, minus the work of friction
... m * g * h - 520 * h
... E = (88 kg * 9.8 m/s^2 - 520 N) * 5.9 m
his velocity is (from K.E. equation)
... v = √(2 * E / m)
while he is slowing down, his average velocity is
... (v + 0) / 2 = v / 2
his stopping time is
... t = .43 m / (v / 2)
his deceleration is
... a = v / t
... m * g * h - 520 * h
... E = (88 kg * 9.8 m/s^2 - 520 N) * 5.9 m
his velocity is (from K.E. equation)
... v = √(2 * E / m)
while he is slowing down, his average velocity is
... (v + 0) / 2 = v / 2
his stopping time is
... t = .43 m / (v / 2)
his deceleration is
... a = v / t
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.