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A wire is stretched 2 mm by a force of 250 N. Determine the force that would stretch the wire 5 mm, assuming that the limit of proportionality is not exceeded.

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Hooke’s law states that extension x is proportional to force F, provided that the limit of proportionality is not exceeded, i.e. x ∝ F or x = kF where k is a constant.

When x = 2 mm, F = 250 N, thus 2 = k(250),

from which, constant k = 2/250 = 1/125

i.e. 5 = (1/125)F

from which, force F = 5(125) = 625 N

Thus to stretch the wire 5 mm a force of 625 N is required.

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