Answer to Question #7256 in Calculus for LeighAnne Manwiller

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 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?

Answer 1

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'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==> k = F/x₁ = 200/16 = 25/2.

Let's find the work needed to stretch a spring from 31 cm to 46 cm:

x₂ = L₃-L₁ = 46 - 31 = 15 cm

A₂ = k*x₂²/2 = 25/2 * 15² / 2 = 5625/4 J.

Let's find the work needed to stretch a spring from 31 cm to 70 cm:

x₃ = L₄-L₁ = 70 - 31 = 39 cm

A₃ = k*x₃²/2 = 25/2 * 29² / 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.