Recent content by blessedcurse

It's being stretched 4 feet beyond its natural length of 2 feet, so the natural length is the 0 and then 4 feet beyond that is 4. At first I had 2 to 6, but then I realized that that meant stretching it from 2 feet beyond its natural length to 6 feet beyond it because the limits refer to change...

I fixed my previous post before you had posted, I thought... sorry! I don't know... I read it as stretching it from 2 to 5 feet beyond its natural length...

Homework Statement The amount of WORK to stretch a spring 4 feet beyond its natural length of 2 feet is 10 ft-lbs. Find the work required to stretch the spring from 4 feet to 7 feet. Homework Equations W=\int^{b}_{a}Fdx=\int^{b}_{a}kxdx=[kx^{2}/2]^{b}_{a} The Attempt at a Solution...

When the spring is at it's natural position of 2 feet, x=0, so F is 0 and W is 0... Is that what you're asking?

Homework Statement The amount of WORK to stretch a spring 4 feet beyond its natural length of 2 feet is 10 ft-lbs. Find the work required to stretch the spring from 4 feet to 7 feet. Homework Equations W=\int^{b}_{a}Fdx=\int^{b}_{a}kxdx=[kx^{2}/2]^{b}_{a} The Attempt at a Solution...

Homework Statement What is the value of xln(x)-x when x=0?Homework Equations I'm assuming you do L'Hopital'sThe Attempt at a Solution I'm assuming you factor out the x, leaving: x(ln(x)-1) but that's still not in the form of \frac{\infty}{\infty} or \frac{0}{0} Would you do...