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Published byΟινώνη Βιτάλης Modified over 5 years ago
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F = k x Springs Web Link: Introduction to Springs
the force required to stretch it the change in its length F = k x k=the spring constant
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Does a larger k value mean that a spring is
easier to stretch, or harder to stretch?
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How far does it stretch if you suspend a 2 N weight from it instead?
Ex: If a spring stretches by 20 cm when you pull horizontally on it with a force of 2 N, what is its spring constant? 2 N How far does it stretch if you suspend a 2 N weight from it instead? 2 N
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F = k x The same equation works for compression:
the force required to compress it the decrease in its length * For an ideal spring, the spring constant is the same for stretching and compressing.
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When a force is exerted on a spring it will either compress (push the spring together) or stretch the spring if the weight is hung on it.
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Some objects like bridges will also behave like springs
Some objects like bridges will also behave like springs. When a weight is placed on a bridge parts will be stretched and under tension, other parts will be be squashed together or compressed
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When the weight and the upward, restoring force are equal the spring is said to be in equilibrium
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Because springs stretch proportionally we can use them as a spring balance to measure a force.
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Determine the spring constant (k) for a single spring by finding the gradient from a graph of F (N) vs x (m) Use masses 50g to 250g, let g = 10ms-2 Repeat for: 2 springs in series 2 springs in parallel Record all data in a labelled table Plot all your data onto one graph (3 lines!) Compare your experimental values for kseries and kparallel with the theoretical formula given below Hookes Law lab
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See Wikipedia for theory!
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Spring Potential Energy, Ep
x F = kx F x kx Work
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ELASTIC FORCE The force Fs applied to a spring to stretch it or to compress it an amount x is directly proportional to x. Fs = - k x Units: Newtons (N) Where k is a constant called the spring constant and is a measure of the stiffness of the particular spring. The spring itself exerts a force in the opposite direction:
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This force is sometimes called restoring force because the spring exerts its force in the direction opposite to the displacement. This equation is known as the spring equation or Hooke’s Law.
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The elastic potential energy is given by:
PEs = ½ kx2 Units: Joules (J)
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5.7 A dart of mass kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its normal length, what speed does the dart acquire? m = 0.1 kg k = 250 N/m x = 0.06 m PEs = K ½ kx2 = ½ mv2 = 3 m/s
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Energy Transformations
Work done by a spring v m m k x
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Energy Transformations
Work done by friction v = 0 m m k x d
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