11.2 - springs

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54 Terms

1
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what does hooke’s law state?

  • force needed to stretch a spring ∝ extension of the spring from its natural length

  • F = k Δ L

2
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force needed to stretch a spring ∝ ?

force needed to stretch a spring ∝ extension of the spring from its natural length

3
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extension of the spring from its natural length ∝ ?

extension of the spring from its natural length ∝ force needed to stretch a spring

4
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what is the equation for hooke’s law?

F = k Δ L

  • F = force needed to stretch a spring

  • k = spring (/ stiffness) constant

  • Δ L = extension from natural length L

5
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F = k Δ L

hooke’s law

6
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what is tension?

  • a pull on the object attached to the string

  • equal and opposite to the force needed to stretch the spring

7
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how can we prove hooke’s law?

  • using a stretched spring supporting a weight at rest

  • by increasing the masses, and therefore tension, we can plot the force against tension

  • hooke’s law states force needed to stretch a spring ∝ extension of the spring from its natural length

  • therefore a force-time graph for hooke’s law will be a straight line through the origin

here

8
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what will the force-time graph for hooke’s law look like?

straight line through the origin

9
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how does spring stiffness vary with k?

the greater the value of k, the more stiff the spring is

10
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what is the unit of k?

N m-1 (newtons per metre)

11
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what is the value of k?

constant for the spring

12
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is the value of k constant?

yes, for the spring

13
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what happens when a spring is stretched beyond its elastic limit?

it doesn’t regain its initial length when the applied force is removed, i.e., the point the spring is plastically / elastically deformed

14
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what is the elastic limit?

  • the point the spring doesn’t regain its initial length when the applied force is removed

  • the point the spring is plastically / elastically deformed

15
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when is a spring elastically deformed?

when the spring passes its elastic limit

16
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springs in parallel picture here

springs in parallel

17
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where is the spring extension for springs in parallel?

here

18
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where is the weight positioned for springs in parallel?

adjusted along the rod to make extension of both springs the same

here

19
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what is the spring extension (Δ L) for a spring in parallel?

the same for each spring, because the weight is adjusted until the extension is the same

20
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for springs in parallel, what is the force needed to stretch each spring?

here

here

FP = kP Δ L

FQ = kQ Δ L

21
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how do you find the weight held up by springs in parallel?

W = k Δ L

  • W = weight

  • k = kP + kQ

22
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for springs in parallel, what is the tension in each spring?

half the weight

23
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for springs in parallel, why was the tension in each spring half the weight?

the weight is supported by both springs, and as the tension is equal and opposite to the weight, the tension in each spring is half the weight

24
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W = k Δ L

weight held up by springs in parallel

25
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derive W = k Δ L

  • weight is supported by both springs

  • W = FP + FQ

  • FP = kP Δ L

    FQ = kQ Δ L

  • therefore W = kP Δ L + kQ Δ L

  • k = kP + kQ

  • therefore W = k Δ L

26
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k + kP + kQ

spring constant for springs in parallel

27
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what is the spring constant for springs in parallel?

k + kP + kQ

28
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springs in series here

springs in series

29
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where is the spring extension for springs in series?

here

30
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for springs in series, what is the tension in each spring?

tension in each spring is equal to each other, and equal to the weight W

31
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for springs in series, is the tension in the each spring each or different to each other?

equal

32
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for springs in series, is the tension equal or different to the weight?

equal

33
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for springs in series, why is the tension in each spring equal to the weight?

the tension is equal and opposite to the force needed to stretch the spring, therefore tension in each spring is equal to weight

34
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for springs in series, what is the extension of each spring?

different to each other

  • Δ LP = W / kP

  • Δ LQ = W / kQ

35
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what is the total extension for springs in series?

  • Δ L = W / k

  • Δ L = Δ LP + Δ LQ

36
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Δ L = Δ LP + Δ LQ

total extension for springs in series

37
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Δ L = W / k

total extension for springs in series

38
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derive Δ L = W / k

  • Δ LP = W / kP

    Δ LQ = W / kQ

  • Δ L = Δ LP + Δ LQ

  • so Δ L = W / kP + W / kQ

  • therefore Δ L = W / k

39
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what is the spring constant for springs in series?

1 / k = 1 / kP + 1 / kQ

40
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1 / k = 1 / kP + 1 / kQ

spring constant for springs in series

41
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derive 1 / k = 1 / kP + 1 / kQ

  • Δ L = W / k

  • = W / kP + W / kQ

  • therefore 1 / k = 1 / kP + 1 / kQ

42
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do the general information thing for each spring type

43
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what energy type is stored in a stretched spring?

elastic potential energy

44
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what is the energy transfer in a stretched spring released from taut?

elastic potential energy = spring’s kinetic energy

45
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what is the work done to stretch the spring by extension Δ L?

W = ½ F Δ L

  • W = work done

  • F = force needed to stretch the spring to extension Δ L

46
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what is the work done stores as in a spring?

elastic potential energy

47
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W = ½ F Δ L

work done to stretch the spring by extension Δ L

48
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what is the elastic potential in a stretched spring?

  • the area under a force-time graph, Ep = ½ F Δ L

  • Ep = ½ k Δ L2

49
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what is the area under a force-time graph for a stretched spring following hooke’s law?

elastic potential energy stores in a stretched spring

50
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where is the elastic potential energy on a force-time graph for a stretched spring following hooke’s law?

area under the graph

51
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Ep = ½ F Δ L

elastic potential energy stores in a stretched spring

52
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derive Ep = ½ F Δ L

  • the work done to stretch a spring to Δ L is stored as elastic potential energy, therefore Ep = ½ F Δ L

  • as the elastic potential energy is the area stored under the graph, which is in a triangle shape, area = 1/2 x base x height, therefore Ep = ½ F Δ L

53
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Ep = ½ k Δ L2

elastic potential energy stores in a stretched spring

54
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derive Ep = ½ k Δ L2

  • the work done to stretch a spring to Δ L is stored as elastic potential energy, therefore Ep = ½ F Δ L

  • hooke’s law states F = k Δ L

  • Ep = ½ (K Δ L) Δ L

  • therefore Ep = ½ k Δ L2