Problem of conservative and nonconservative forcesby dmatador Tags: conservative, forces, nonconservative 

#1
Oct2210, 07:28 PM

P: 122

A 65.1 kg person jumps from rest off a 3.04 mhigh tower straight down into the water. Neglect air resistance during the descent. She comes to rest 1.11 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.
I've tried to solve the first part using that fact that the initial energy and the final one right before entering the water will be equal, then finding the final velocity and using this to find work and then force, but it doesn't work. I need some help. 



#2
Oct2210, 07:48 PM

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P: 5,966

Please show how you calculated initial and final energies so we can see where you may have gone wrong. You don't have to find the 'final' energy at the water surface (although it's OK to do so), but you can find the final energy at the rest point below the water surface, and save a step.




#3
Oct2210, 08:00 PM

P: 122

sorry about the notation...
(m)(g)(h_f) + (.5)(m)(v_f)^2 = (m)(g)(h_i) + (.5)(m)(v_i)^2 I then eliminated the m's, the terms (m)(g)(h_f) and (.5)(m)(v_i)^2 because the final height is zero and the initial velocity is zero. I then just solved for v_f, the final velocity. Then I used W = (m)(g)(h_f) + (.5)(m)(v_f)^2  (m)(g)(h_i)  (.5)(m)(v_i)^2 This is because once the swimmer hits the water, there is a change in total energy. I used the final velocity, v_f, from the other equation as v_i in this one. This is because she hits the water with this speed. I then used 0 as initial height h_i to eliminate the term (m)(g)(h_i) and v_f = 0 to eliminate (.5)(m)(v_i)^2. I then solved for W and divided this by the distance 1.11 done by this work to get the force. 



#4
Oct2210, 08:17 PM

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Problem of conservative and nonconservative forces
You seem to be on the right track, but you are not showing your calcs. You may have made a math error, or perhaps you slipped up on a minus sign when calculating the potential energy at the 'rest' position.



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