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?

Question

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?

anonymous · Answer

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!

anonymous · Answer

Thank you, but the problem wants the work, not the force...

anonymous · Answer

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,$$W=\frac{11000}{3}((.60-.33)^2-(.44-.33)^2)J=222\frac{2}{15}J$$

anonymous · Answer

Sorry, I apparently didn't read carefully!