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Work of 4 Joules is done in stretching a spring from its natural length to 18 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (18 cm)?
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First, note that work = fd We know that force is a function of its distance d from equilibrium, f(d) = kd, d is in meters. W = (1/2)kd^2. We are given that 4 joules of work was used to stretch the string a distance of 18 cm = 0.18 meters: W = 4 4 = (1/2)k(0.18)^2 So, we know everything in that equation except k, the spring constant. Solve for k, then, once we know k, recall that f(d) = kd.
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