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If 196 ft-lb of work are needed to stretch a spring 3.5 ft beyond equilibrium, how far will the spring stretch if a 17-lb weight is attached to its end? (Use decimal notation. Give your answer to two decimal places.)

Answer :

The spring will stretch approximately 0.22 ft when the 17-lb weight is attached.

According to Hooke's law, the force exerted by a spring is directly proportional to its displacement. Mathematically, this can be expressed as F = kx, where F is the force, k is the spring constant, and x is the displacement.

Given that 196 ft-lb of work is needed to stretch the spring 3.5 ft beyond equilibrium, we can calculate the spring constant. The work done on the spring is equal to the potential energy stored in the spring, which can be expressed as W = (1/2)kx², where W is the work done and x is the displacement. Rearranging the equation, we have k = 2W/x².

Substituting the given values, we get k = 2 * 196 ft-lb / (3.5 ft)² = 112 lb/ft.

Now, we can use Hooke's law to find the displacement when the 17-lb weight is attached. Rearranging F = kx, we have x = F/k. Plugging in the values, x = 17 lb / 112 lb/ft ≈ 0.152 ft or 0.15 ft (rounded to two decimal places). Therefore, the spring will stretch approximately 0.15 ft or 0.22 ft (rounded to two decimal places) when the 17-lb weight is attached.

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