## Related questions with answers

If 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another 10 J is needed to stretch it from 12 cm to 14 cm, what is the natural length of the spring?

Solutions

VerifiedLet the natural length be $l$

Then we can write

$6=\int_{10-l}^{12-l}kx\hspace{1mm}dx$

$6= \bigg[\dfrac{1}{2}kx^2\bigg]_{10-l}^{12-l}$

$6= \dfrac{1}{2}k\bigg[\left( 12-l\right)^2-\left( 10-l\right)^2\bigg]\rightarrow\text{\color{#c34632}Equation 1}$

Applying Hooke's law, the force required to keep a spring stretched $x$ units more than its natural length is proportional to $x$:

$f(x)=k x$

where k is a positive constant (called the spring constant).

Obtaining that the work done by the force $f (x)$, to move an object from $x = a$ to $x = b$ is calculated as:

$W = \int_ {a} ^ {b} k x d x \tag {1}$

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