# Answer to Question #7256 in Calculus for LeighAnne Manwiller

Question #7256
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 31 cm. If a 200 Newton force is required to keep the spring stretched to a length of 47 cm, how much work is required to stretch it from 46 cm to 70 cm?
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 31 cm. If a 200 Newton force is required to keep the spring stretched to a length of 47 cm, how much work is required to stretch it from 46 cm to 70 cm?

Let&#039;s make the following denominations:
L1 = 31 cm (natural length)
L2 = 47 cm
F = 200 N
A1 = 2 J
L3 = 46 cm
L4 = 70 cm

We need to find the stiffness of the spring:
x1 = L2-L1 = 47 - 31 = 16 cm;
F = k*x1 ==&gt; k = F/x1 = 200/16 = 25/2.

Let&#039;s find the work needed to stretch a spring from 31 cm to 46 cm:
x2 = L3-L1 = 46 - 31 = 15 cm
A2 = k*x2&sup2;/2 = 25/2 * 15&sup2; / 2 = 5625/4 J.

Let&#039;s find the work needed to stretch a spring from 31 cm to 70 cm:
x3 = L4-L1 = 70 - 31 = 39 cm
A3 = k*x3&sup2;/2 = 25/2 * 29&sup2; / 2 = 21025/4 J.

So, the work needed to stretch a spring from 60 cm to 74 cm is
A = A3 - A2 = 21025/4 - 5625/4 = 3850 J.

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