Question

A spring has a natural length of 7 m. If a 4-N force is required to keep it stretched to a length of 11 m, how much work W is required to stretch it from 7 m to 13 m?

Answer 1

Answer:

18 J is the work required to stretch a spring from 7 m to 13 m.

Step-by-step explanation:

The work done is defined to be the product of the force $F$ and the distance $d$  that the object moves:

$W=Fd$

If $F$ is measured in newtons and $d$ in meters, then the unit for is a newton-meter, which is called a joule (J).

This definition work as long as the force is constant, but if the force is variable like in this case, we have that the work done is given by

$W=\int\limits^b_a {f(x)} \, dx$

Hooke’s Law states that the force required to maintain a spring stretched $x$    units beyond its natural length is proportional to

$f(x)=kx$

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

To find how much work W is required to stretch it from 7 m to 13 m you must:

Step 1: Find the spring constant

We know that the spring has a natural length of 7 m and a 4 N force is required to keep it stretched to a length of 11 m. So, applying Hooke’s Law

$4=k(11-7)\\\\\frac{k\left(11-7\right)}{4}=\frac{4}{4}\\\\k=1$

Thus $F=x$

Step 2: Find the the work done in stretching the spring from 7 m to 13 m.

Recall that the natural length is 7 m, so when we stretch the spring from 7 m to 13 m, we are stretching it by 6 m beyond its natural length.

Work needed to stretch it by 6 m beyond its natural length

$W=\int\limits^6_0 {x} \, dx \\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\left[\frac{x^{1+1}}{1+1}\right]^6_0\\\\\left[\frac{x^2}{2}\right]^6_0=18$

18 J is the work required to stretch a spring from 7 m to 13 m.