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    in this problem you will study two cases of springs connected in parallel that will enable you to draw a general conclusion.

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    MasteringPhysics 2.0: Problem Print View

    Part A

    What is the effective spring constant of the two-spring system?

    Hint A.1 Free-body diagram

    Draw a free-body diagram for spring 2. Notice that the system is in equilibrium. The free-body diagram should help you to

    find the force, acting to the right, on spring 1, and

    find the extension of spring 2, in terms of known quantities (including ).

    Hint A.2 Free-body diagram for spring 2

    Part A.3 Determine the extension of spring 2

    What is the extension of spring 2?

    Give your answer in terms of and

    ANSWER: =  F/k_2

    Part A.4 Determine the extension of spring 1

    Part A.5 Determine the total extension of the two springs

    Part A.6

    Part A.7 Determine the extension of the equivalent system

    Hint A.8 Solving for

    Express the effective spring constant in terms of and .

    ANSWER:

    =  (1/k_1+1/k_2)^(-1)

    Source : notendur.hi.is

    Solved Question 7 In this problem you will study two cases

    Answer to Solved Question 7 In this problem you will study two cases

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    Question: Question 7 In This Problem You Will Study Two Cases Of Springs Connected In Parallel That Will Enable You To Draw General Conclusion Figure 1 Of 2 Spring 1,K Wwwww Wwww Spring 2, K2 F Www Figure 2 Of 2 Spring 1, K Mwwи Spring 2, K2 F Www Spring 3, K3 Www Two Springs In Parallel Consider Two Massless Springs Connected In Parallel. Springs 1 And 2 Have

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    Transcribed image text: Question 7 In this problem you will study two cases of springs connected in parallel that will enable you to draw general conclusion Figure 1 of 2 spring 1,k wwwww wwww spring 2, k2 F Www Figure 2 of 2 spring 1, k mwwи spring 2, k2 F www spring 3, k3 Www Two springs in parallel Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thin, vertical rod. A constant force of magnitude F is being exerted on the rod. The rod remains perpendicular to the direction of the applied force, so that the springs are extended by the same amount. This system of two springs is equivalent to a single spring, of spring constant k. (Figure 1) Part A Find the effective spring constant k of the two-spring system. Give your answer for the effective spring constant in terms of k1 and k2. View Available Hint(s) ΑΣφ k Three springs in parallel Now consider three springs connected in parallel as shown. (Figure 2)The spring constants of springs 1, 2, and 3 are k1, k2, and k3. The springs are connected by a vertical rod, and a force of magnitude F is being exerted to the right. Part B Find the effective spring constant k' of the three-spring system. Give your answer in terms of k1, k2, and k3. ? ΑΣφ

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    What is the spring constant in parallel connection and series connection?

    Parallel. When two massless springs following Hooke's Law, are connected via a thin, vertical rod as shown in the figure below, these are said to be connected in parallel. Spring 1 and 2 have spring constants k_1 and k_2 respectively. A constant force vecF is exerted on the rod so that remains perpendicular to the direction of the force. So that the springs are extended by the same amount. Alternatively, the direction of force could be reversed so that the springs are compressed. This system of two parallel springs is equivalent to a single Hookean spring, of spring constant k. The value of k can be found from the formula that applies to capacitors connected in parallel in an electrical circuit. k=k_1+k_2 Series. When same springs are connected as shown in the figure below, these are said to be connected in series. A constant force vecF is applied on spring 2. So that the springs are extended and the total extension of the combination is the sum of elongation of each spring. Alternatively, the direction of force could be reversed so that the springs are compressed. This system of two springs in series is equivalent to a single spring, of spring constant k. The value of k can be found from the formula that applies to capacitors connected in series in an electrical circuit. For spring 1, from Hooke's Law F=k_1x_1 where x_1 is the deformation of spring. Similarly if x_2 is the deformation of spring 2 we have F=k_2x_2 Total deformation of the system x_1+x_2=F/k_1+F/k_2 =>x_1+x_2=F(1/k_1+1/k_2) Rewriting and comparing with Hooke's law we get k=(1/k_1+1/k_2)^-1

    What is the spring constant in parallel connection and series connection?

    Physics Waves and Vibrations Simple Harmonic Motion - Springs

    1 Answer

    A08 · Stefan V. Dec 15, 2016

    Parallel.

    When two massless springs following Hooke's Law, are connected via a thin, vertical rod as shown in the figure below, these are said to be connected in parallel. Spring 1 and 2 have spring constants

    k 1 and k 2

    respectively. A constant force

    → F

    is exerted on the rod so that remains perpendicular to the direction of the force. So that the springs are extended by the same amount. Alternatively, the direction of force could be reversed so that the springs are compressed.

    http://www.chegg.com/homework-help/questions-and-answers

    This system of two parallel springs is equivalent to a single Hookean spring, of spring constant

    k . The value of k

    can be found from the formula that applies to capacitors connected in parallel in an electrical circuit.

    k = k 1 + k 2

    Series.

    When same springs are connected as shown in the figure below, these are said to be connected in series. A constant force

    → F

    is applied on spring 2. So that the springs are extended and the total extension of the combination is the sum of elongation of each spring. Alternatively, the direction of force could be reversed so that the springs are compressed.

    notendur.hi.is/eme1/skoli/edl_h05/masteringphysics

    This system of two springs in series is equivalent to a single spring, of spring constant

    k . The value of k

    can be found from the formula that applies to capacitors connected in series in an electrical circuit.

    For spring 1, from Hooke's Law

    F = k 1 x 1 where x 1

    is the deformation of spring.

    Similarly if x 2

    is the deformation of spring 2 we have

    F = k 2 x 2

    Total deformation of the system

    x 1 + x 2 = F k 1 + F k 2 ⇒ x 1 + x 2 = F ( 1 k 1 + 1 k 2 )

    Rewriting and comparing with Hooke's law we get

    k = ( 1 k 1 + 1 k 2 ) − 1 Answer link

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