Physics Help Forum Potential energy confusion

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 May 19th 2017, 07:55 AM #1 Member   Join Date: Jan 2015 Posts: 69 Potential energy confusion Hi everyone, I have a Gravitational Potential energy problem that I thought was straightforward but the solution to the problem is not what I worked out and I don't understand why ? the problem is as follows - Problem A uniform 3.00 kg rope, 24.0 m long lies on the ground at the top of a vertical cliff. A mountain climber at the top lets down half of it to his partner to help him climb the cliff. What is the change in Potential energy of the rope during this maneuver ? Now I thought that Gravitational Potential energy is given by .. U = m g h , where U is potential energy m is mass and h the height of the mass This being so then I calculated that the change in PE is U(1) -U(2) = m g h(1) - 1/2 m g h(2) , where h(1) is height of rope at top of the cliff and h(2) is is height of rope at a height 12.0 metres lower that is 1/2 length of rope lower so .. U(1) - U(2) = 1/2 m g (h(1) - h(2)) that is .. U(1) - U(2) = 1/2 m g (-12.0) = 1.5 x 9.8 x(-12.0) = (-176.4J) However the answer in the book is half this value , namely (-88.2J) ? I don't understand why this is the correct solution .. can anyone tell me why my solution is wrong ? Jackthehat
May 19th 2017, 08:23 AM   #2
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 A uniform 3.00 kg rope, 24.0 m long lies on the ground at the top of a vertical cliff. A mountain climber at the top lets down half of it to his partner to help him climb the cliff. What is the change in Potential energy of the rope during this maneuver ?
Let the cliff be $h = 12.0 \, m$ high to make the calculation simple.

$U_0 = mgh$

half the rope is at $h = 12 m$ and the center of mass of the other half is at $\dfrac{h}{2} = 6 \, m$

$U_f = \dfrac{m}{2}gh + \dfrac{m}{2}g \cdot \dfrac{h}{2} = \dfrac{3mgh}{4}$

$\Delta U = U_f - U_0 = \dfrac{3mgh}{4} - mgh = -\dfrac{mgh}{4} = -88.2 \, J$

 May 19th 2017, 08:34 AM #3 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,171 Skeeter is using a coordinate system where h = 0 is at the base of the cliff, whereas Jack is using a coordinate system where h=0 is at the top of the cliff. Either approach is acceptable (as long as we're consistent). So using Jack's coordinate system: U_1 = 0 U_2 = (-12/2 m) x (3/2 Kg) x (9.8 m/s^2) = -88.2 J Note that the distance traveled by the center of mass of the portion of rope dropped over the side of the cliff is half the height of the cliff. U_2 - U_1 = -88.2 J - 0 J = -88.2 J
 May 19th 2017, 09:46 AM #4 Member   Join Date: Jan 2015 Posts: 69 Thank you for taking the time to explain the solution to the problem. I understand where I went wrong. I had taken the full length of the rope as 'h' in the equation U = m g h... instead of taking the height of the centre of mass of the string. Your help was much appreciated. Regards, Jackthehat
 May 19th 2017, 09:47 AM #5 Member   Join Date: Jan 2015 Posts: 69 Hi ChipB, Thank you for taking the time to explain the solution to the problem. I understand where I went wrong. I had taken the full length of the rope as 'h' in the equation U = m g h... instead of taking the height of the centre of mass of the string. Your help was much appreciated. Regards, Jackthehat
 May 19th 2017, 06:28 PM #6 Senior Member   Join Date: Aug 2010 Posts: 150 As long as it is a "linear" change, as here, the calculation can be done using the "average" distance. That is, treating the mass as if it were concentrated half way done the cliff. That is the reason the correct answer is 1/2 your answer where you are, essentially, treating the mass as if it were at the top of the cliff.
 May 20th 2017, 09:29 AM #7 Senior Member   Join Date: Jun 2010 Location: NC Posts: 341 Rope from a Cliff - interesting Problem A uniform 3.00 kg rope, 24.0 m long lies on the ground at the top of a vertical cliff. A mountain climber at the top lets down half of it to his partner to help him climb the cliff. What is the change in Potential energy of the rope during this maneuver ? Solution: Skeeter is using a coordinate system where h = 0 is at the base of the cliff, whereas Jack is using a coordinate system where h=0 is at the top of the cliff. Either approach is acceptable (as long as we're consistent). So using Jack's coordinate system: U_1 = 0 U_2 = (-12/2 m) x (3/2 Kg) x (9.8 m/s^2) = -88.2 J Note that the distance traveled by the center of mass of the portion of rope dropped over the side of the cliff is half the height of the cliff. U_2 - U_1 = -88.2 J - 0 J = -88.2 J ################### Coment and Further Question: It seems the rope is modeled as an "extended BODY" with its "shape" as a charscteristic (not property) and its PE represented in terms of elevation-of center-of-mass of its discrete parts. With the lowering event, the PE change, (U2 - U1) , was calculated to be -88.2J. How does this reconcile with the principle of Conservation of Energy? Does the "-" sign imply energy transfer "across the rope boundary" or "out of" the rope? If the energy of the rope decreased, did the energy of something else increase. What something else? What happens here? Thanks JP Last edited by THERMO Spoken Here; Yesterday at 12:46 PM. Reason: rephrasing

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