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A wire is stretched 1.5 mm by a force of 300 N. Determine the force that would stretch the wire 4 mm, assuming the elastic limit of the wire is not exceeded. |
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Answer» Hooke's law states that extension x is proportional to force F, provided that the limit of proportionality is not exceeded, i.e. x α F or x = kF where k is a constant. When x = 1.5 mm, F = 300 N, thus 1.5 = k(300), from which, constant k = \(\cfrac{1.5}{300}\) = 0..005 When x = 4 mm, then 4 = kF i.e. 4 = 0.005 F from which, force F = \(\cfrac{4}{0.005}\) = 800 N Thus to stretch the wire 4 mm, a force of 800 N is required. |
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