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
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.
1 answer
1 answer