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Question
The length of nylon rope from which a mountain climber is suspended has a force constant of $1.40 \times 10^4 \textrm{ N/m}$. (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy. (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
  1. $1.99 \textrm{ Hz}$
  2. $56.9 \textrm{ cm}$
  3. $1.40 \textrm{ Hz}$, $84.8 \textrm{ cm}$
Solution Video

OpenStax College Physics Solution, Chapter 16, Problem 35 (Problems & Exercises) (7:09)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A mountain climber is hanging from a nylon rope and the frequency with which they will bounce on the rope if they fall a little bit is one over two pi, times the square root of the spring constant of the rope, divided by the total mass of the climber plus his equipment. So that's one over two pi times square root of 1.4 times ten to the four Newtons per meter divided by 90 kilograms which is 1.99 Hertz. In part B it asks how much would this rope stretch to prevent the climber from falling if he free-falls two meters. So he begins here, say, at this top blue dotted line and then free-falls two meters and then reaches this position here when the rope just begins to stretch. Then the rope will stretch all the way until this point here when he stops falling and we'll call this the reference level y equals zero. So we don't know what this height is here, this distance above zero during which the rope stretches. But we do know that between here and here when the rope is slack, we're told that's two meters. So the y position that he starts with, given that we've chosen the reference level to be y equals zero at the very bottom, this top y position is going to be h, the amount of stretching plus two meters. So we're going to use conservation of energy to figure out what this height is going to be and we know that the gravitational potential energy that the climber has at this very initial position is going to equal the elastic potential energy that they have at this position. At this position there will be zero gravitational potential energy because this is y equals zero, we've chosen this to be the reference level. So we can equate these two things, gravitational potential energy is going to be m g times h plus two meters and that's going to equal the elastic potential energy that the climber has at this reference level y equals zero. The elastic potential energy will be one half times the spring constant times the amount by which the nylon rope is stretched which is an amount h and we square that. So we distribute the m ginto the brackets here and then take both terms to the right hand side and then switch the sides around. We end up with one half k h squared minus m g h, minus two m g equals zero. This is a quadratic equation that we need the quadratic formula to solve.  So we have one half times the spring constant times h squared, minus the mass times acceleration due to gravity times h, minus two times m g and that equals zero. This works out to 7000 h squared minus 882.9 h minus 1765.8 equals zero. Then we substitute that into the quadratic formula  where we have the coefficient of the linear term written here with the opposite sign. You might be more familiar with this form, so x equals negative b plus or minus b squared minus four a c, all over two a, where b is the coefficient of linear term, a is the coefficient of the squared term and c is the constant term. Okay. So plugging all this into our calculator and choosing a plus here because that's going to give a positive answer. In choosing negative we get a negative answer here which would have no physical meaning and so we take the positive there. We end up with 0.569 which is 56.9 centimeters. That's the amount by which the rope will stretch such that the climber will have all of their initial gravitational potential energy turned into elastic potential energy at the reference level. Now in part C we're told that suppose you have a nylon rope that is twice the length of the one used in parts A and B. Answer the questions again. What will the frequency of the climber's bouncing be and what will the stretch of the rope be in this case? We're going to model this new rope as two ropes in series or two springs in series. We'll have to figure out what is the effective spring constant of this combination. So the force that the rope is going to apply upwards is going to be some equivalent spring constant multiplied by x plus x where x is the amount by which the first rope stretches. The second rope is going to stretch by the same amount because each segment here has the same spring constant and so we would expect them to stretch by the same amount. That's going to be two k eq x. Now this force that the rope exerts upwards is going to be the same as the force exerted by each segment and so we equate those here. That's going to be two k eq x then is going to be k x because the force exerted by the first segment is the spring constant that we know for this segment, times the amount by which the particular segment stretches. The equivalent spring constant then is k over two. The x's cancel and divide both sides by two. We answer all of our questions now using k over two as our spring constant. So the frequency of bouncing is one over two pi times square root of this equivalent spring constant over mass. Substituting k over two in place of that, we get one over two pi times square root k over two m. So that's one over two pi times 1.4 times ten to the four Newtons per meter over two times 90 kilograms giving us 1.40 Hertz. Then the height, we're going to use this here. The only thing that's changed is the spring constant k so we're going to replace k with 1.4 times ten to the four divided by two, that's what we did here, and then the other terms are going to be the same because they're not dependent on k. They just have m g in both of them. We can copy those numbers down. So we have 3500 h squared minus 882.9 h minus 1765.8 equals zero. Putting that into the quadratic formula gives us 84.8 centimeters is the amount that the rope will stretch when two of them are connected in series.