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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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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!

  2. anonymous
    • 5 years ago
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    Thank you, but the problem wants the work, not the force...

  3. anonymous
    • 5 years ago
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    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,\[W=\frac{11000}{3}((.60-.33)^2-(.44-.33)^2)J=222\frac{2}{15}J\]

  4. anonymous
    • 5 years ago
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    Sorry, I apparently didn't read carefully!

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