The "spring-flex" exercise system consists of a spring with one end fixed and a handle on the other end. The idea is that you exercise your muscles by stretching the spring from its natural length, which is 33 cm. If a 110 Newton force is required to keep the spring stretched to a length of 46 cm, how much work is required to stretch it from 44 cm to 60 cm?

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The "spring-flex" exercise system consists of a spring with one end fixed and a handle on the other end. The idea is that you exercise your muscles by stretching the spring from its natural length, which is 33 cm. If a 110 Newton force is required to keep the spring stretched to a length of 46 cm, how much work is required to stretch it from 44 cm to 60 cm?

Mathematics
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F=-kx, so 110=(-k)(46-33) 110/13=-k So the force required to stretch it from 44 to 60cm is the force required to stretch it to 60 minus the force required to stretch it to 44 F=(-k)60-(-k)44=(-k)(60-44)=(-k)16=(110/13)16=135.4N Remember your significant figures on your final answer!
Thank you, but the problem wants the work, not the force...
You need to work out the spring constant first. You're told the equilibrium position is 33cm, and upon stretching to 46cm, the force required is 110N. Hence,\[F=-k(x-x_0) \rightarrow =110N=-k(0.46m-.33m)\rightarrow k=-\frac{11000}{13}N/m\]The work is given by,\[W=\int\limits_{x_i}^{x_f}F.dx=\int\limits_{x_i}^{x_f}(-k(x-x_0)).dx\]\[=-\frac{k}{2}((x_f-x_0)^2-(x_i-x_0)^2)\]Inserting your values,\[W=\frac{11000}{3}((.60-.33)^2-(.44-.33)^2)J=222\frac{2}{15}J\]

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Sorry, I apparently didn't read carefully!

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