Question

A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of 60 cm. Find the work done in stretching the spring from 60 cm to 65 cm. First, setup an integral and find a a, b b, and f ( x ) f(x) which would compute the amount of work done.

Answer 1

Answer:

The work done in stretching the spring is 0.875 J.

Explanation:

Given that,

Force = 140 N

Natural length = 60-40 = 20 cm

Stretch length of the spring = 65-60 = 5 cm

We need to calculate the spring constant

Using formula of Hooke's law

$F= kx$

$140=k\times20\times10^{-2}$

$k=\dfrac{140}{20\times10^{-2}}$

$k=700$

We need to calculate the work done

$W=\int_{a}^{b}{kx}dx$

$=\int_{0}^{0.05}{700x}dx$

On integration

$W=700\times(\dfrac{x^2}{2})_{0}^{0.05}$

$W=700\times(\dfrac{(0.05)^2}{2}-0)$

$W=0.875\ J$

Hence, The work done in stretching the spring is 0.875 J.