Work - Area under curve of F vs. d graph

[jtw2e]

Homework Statement

Homework Equations

W=FD
Area under curve = Work

The Attempt at a Solution

Note, although the problem doesn't say this, and the given graph and description are deceptive, Force is NOT constant.

I need to somehow find the Force required to move the cart from 2.5 to 4.2 ft so that I can find the area under the two sections... at least I think that's how to do it. But I have no idea how in the world to find the Force. I've tried work-energy theorem but don't have enough info (don't know v1 or v2 or accel).

Please help me understand this. They did not cover it or even address it in lecture and it was assigned today and is due tonight.

 

[Rayquesto]

As I understand, You must take the integral from 2.5-a to 4.2=b of the equation 28/2.5x and find the anti-derivative of 28/2.5x(11.2x), since for every 11.2 pounds a foot is increased and vice-versa, kind of like the spring constant in a spring if you are familiar with that.
you end up with 1/2 times 11.2 times (4.2 squared minus 2.5 squared) which is about 63.784lb-ft. is that the answer for a?
as far as for b, I think its just 11.2 times (4.2 squared minus 2.5 squared) since the anti-derivative of the 11.2x equation respresnts s(x) aka position and 11.2x itself respresents v(x) velocity. So, the answer for b is probably 127.568m/s. that the answer for b?

 

[Rayquesto]

and when they mean force is not constant, it means that it changes as displacement does, there for you would acquire an equation of the increase of force as displacement increases. so the formula would be 28lb/2.5meters, which is 11.2lb/m or 11.2x.

 

[jtw2e]

and when they mean force is not constant, it means that it changes as displacement does, there for you would acquire an equation of the increase of force as displacement increases. so the formula would be 28lb/2.5meters, which is 11.2lb/m or 11.2x.

As I understand, You must take the integral from 2.5-a to 4.2=b of the equation 28/2.5x and find the anti-derivative of 28/2.5x(11.2x), since for every 11.2 pounds a foot is increased and vice-versa, kind of like the spring constant in a spring if you are familiar with that.
you end up with 1/2 times 11.2 times (4.2 squared minus 2.5 squared) which is about 63.784lb-ft. is that the answer for a?
as far as for b, I think its just 11.2 times (4.2 squared minus 2.5 squared) since the anti-derivative of the 11.2x equation respresnts s(x) aka position and 11.2x itself respresents v(x) velocity. So, the answer for b is probably 127.568m/s. that the answer for b?

Thanks very much for your detailed reply, but we are not covering integrals until next semester; the only concept we're supposed to use is dealing with finding the Area of trapezoids under the curve with regular geometry. And also k = 1/2mv^2, and possibly W_tot=K₂ - K₁. So I don't know the technique to take the integral or find an anti-derivative, and am expected to solve this by a different method. :( I'm sorry, I do not know whether our value for a) and b) are correct.

 

[Rayquesto]

AH darn! That makes things too complicated. riemann sums are really tough. so what you should do instead is just find the area under the curve. so A(x)= the limit as n approaches infinity of the sigma series (i/n)(11.2x) where i is n(nplus 1)/2 and n is the amount of squares ...so A(x)=(11.2x/n)(n^2plusn)/2=11.2xnplus 11.2x/2 and suppose we have 100 squares. at x=2.5 which would be 1414 oh crap I am sorry this isn't right but i got to go. be back in a few minutes.