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Down the Pole. A fireman of mass m slides a distanc down a
Chapter 9, Problem 61P(choose chapter or problem)
Down the Pole. A fireman of mass m ? ? slides a distanc? ? down a pole. He starts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance ?h? ? ?d? above the ground and descended with negligible air resistance. (a) What average friction force did the fireman exert on the pole? Does your answer make sense in the special case of ?h? = ?d? and ?h? = 0? (b) Find a numerical value for the average fiction force 75 kg fireman exerts, for ?d? = 2.5 m and h? ? = 1.0 m. (c) In terms of ?g?, ?h?? ?. what is the speed of the fireman when he is a distance ?y? above the bottom of the pole?
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QUESTION:
Down the Pole. A fireman of mass m ? ? slides a distanc? ? down a pole. He starts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance ?h? ? ?d? above the ground and descended with negligible air resistance. (a) What average friction force did the fireman exert on the pole? Does your answer make sense in the special case of ?h? = ?d? and ?h? = 0? (b) Find a numerical value for the average fiction force 75 kg fireman exerts, for ?d? = 2.5 m and h? ? = 1.0 m. (c) In terms of ?g?, ?h?? ?. what is the speed of the fireman when he is a distance ?y? above the bottom of the pole?
ANSWER:Solution 61P Step 1 of 3: a) Speed at ground if steps off platform at height h K + U + W = K + U 1 1 other 2 2 1 2 2 mgh = mv2, so2v = 2gh2 Motion from top to bottom of pole(y=0 at bottom) K + U + W = K + U 1 1 other 2 2 1 2 mgd fd = mv2 2 We have v = 2ghand substitute in above equation 2 mgd fd = mgh fd = mgd mgh fd = mg(d h) Dividing by d mg(dh) f = d = mg(1 ) h d If h = 0, f = (friction has no effect) If h = 0, v2= 0( no motion) The equation for f gives f=mg in this case .When f=mg the forces on him cancel and doesn’t accelerate down the pole, which agrees with v = 0. 2