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Work done by a Spring Force question

  1. Nov 19, 2008 #1
    A 220 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.8 N/cm (Fig. 7-30). The block becomes attached to the spring and compresses the spring 11 cm before momentarily stopping.

    (a) While the spring is being compressed, what work is done on the block by the gravitational force on it?

    (b) What work is done on the block by the spring force while the spring is being compressed?

    (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

    (d) If the speed at impact is doubled, what is the maximum compression of the spring?



    I know that for the first part W=mg*delta y and the second part should be W=1/2Kx^2 but then i get a little lost on the last two questions. I know this is a spring force and gravitational question but does it also use kinetic? Every time i plug the numbers into the formula for kinetic energy it comes out wrong.



    Thanks!
     
  2. jcsd
  3. Nov 19, 2008 #2

    Hootenanny

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    Welcome to Physics Forums.

    Using the first two equation can you calculate the total work done on the block whilst the spring is being compressed?

    And a hint for the next part: Which quantity is conserved whilst the spring is being compressed?
     
  4. Nov 19, 2008 #3
    So to get the total work i would add the two amounts together.. I'm just a bit confused because it ends up being negative work...
     
  5. Nov 19, 2008 #4

    Hootenanny

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    Hmm... The work done by gravity should be positive since the force (weight) is in the same direction as the displacement, the work done by the spring should be negative since the force of the spring is in the opposite direction to the displacement. However, since in this case the work done by gravity is greater than that done by the spring, the total work done should be positive.

    Would you mind posting your calculations for the first two parts. Note also that the correct form of the expression for the work done by the spring is -1/2kx2 for the reasons I have above.
     
  6. Nov 19, 2008 #5
    a. W=(.22kg)(9.8m/s^2)(.11m)
    W=.23716 J

    b.W=-1/2(.028N/m)(.11^2)
    and then i get W=-1.694e-4...which confuses me because the correct answer should be -1.694 J...
     
  7. Nov 19, 2008 #6

    Hootenanny

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    2.8 N/cm [itex]\neq[/itex] 0.028 N/m.

    Notice that the units are Newton's divided by centimetres, therefore to convert the constant into N/m you should divide by 0.01m.

    Does that make sense?
     
    Last edited: Nov 19, 2008
  8. Nov 19, 2008 #7
    Yeah that does thank you.
    I still don't understand how the total work would be positive then, is this correct
    .237J-1.694J=-1.457
    then would i plug that into the equation for kinetic energy and solve for velocity for part c?
     
  9. Nov 19, 2008 #8

    Hootenanny

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    Whoops, when I initially calculated the work done by the spring, I thought that the spring constant was in N/m. Your answers are correct.

    So, what about the hint I gave in my first post: which quantity is conserved?
     
  10. Nov 19, 2008 #9
    Well the gravitational energy should have been conserved but when i try .237J=1/2(.22kg)v^2 i get 1.467 m/s, which isn't right.
     
  11. Nov 19, 2008 #10

    Hootenanny

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    Hmm... you're on the right tracks, but think bigger. HINT: There are no dissipative forces present.
     
  12. Nov 19, 2008 #11
    Okay the only other force i can think of is that of the earth, so (.22kg)(9.8)=1/2(.22)v^2...which is wrong too.
     
  13. Nov 19, 2008 #12

    Hootenanny

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    I meant that the total energy of the system is conserved. Do you understand why?
     
  14. Nov 19, 2008 #13
    I think so, but if the total energy is negative how do i find the velocity?
     
  15. Nov 19, 2008 #14

    Hootenanny

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    Can you write out an equation that represents the concept of conservation of energy in this case?
     
  16. Nov 19, 2008 #15
    Well i think this is whats confusing me. I know that the intial energy plus the work has to equal the final energy, but i don't know which is which.
     
  17. Nov 19, 2008 #16

    Hootenanny

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    Okay, I'll set this up for you:

    So, when the spring is fully compressed:
    Gravitational PE1 + Elastic PE1 + Kinetic Energy1 = const.

    Equally, when the mass leaves the spring:
    Gravitational PE2 + Elastic PE2 + Kinetic Energy2 = const.

    Subtracting the two equations (and noting that the constants are equal in each case we obtain):

    ΔGravitational PE + ΔElastic PE + ΔKinetic Energy = 0

    Do you follow?
     
  18. Nov 19, 2008 #17
    Okay that makes more sense. So Ek=-.237+1.694. Then from there i plug it into the equation Ek=1/2mv^2 and get that v=3.639 m/s. So if the speed is then doubled, i would solve for a new kinetic energy and then using the conservation of energy find the elastic energy, then solve for the compression. Thanks so much!
     
  19. Nov 20, 2008 #18

    Hootenanny

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    Sounds good to me :approve:
     
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