Calculating Work Done: Spring Force vs. Chain Weight | Integral Setup Help"

[Sheneron]

Homework Statement

A spring exerts 4x pounds of force when stretched x feet from rest. One end is fixed to the ceiling. A chain that is 10ft long and weighs 2lb/ft hangs from the other end. The end of the chain just brushes the floor. Find the work done pulling down on the chain a distance of 2ft.

I am having trouble setting up this integral. The force of the spring would be 4x and you would minus the force of the chain which would be 2(10 - x)? Apparently that isn't right though. If someone could help me set this up it would be appreciated. Thanks

 

[Fredrik]

You're supposed to integrate the force over an interval, and you know the force is 4x (or -4x). What you need to do is figure out the endpoints of the interval. It's obviously (a,a+2) where a must be determined from the condition you stated in the first paragraph.

Yes, you could take the endpoints of the interval to be (0,2), but then the force isn't 4x. Can you figure out what the force is in this case? (The answer involves the same "a").

 

[Dick]

When you start pulling the chain is how far is the end of the spring from the ceiling? The weight of the chain is 20lbs and the force exerted by the spring is 4x. Let x be the distance from the ceiling. What's the range of x? What's the force exerted by the spring at distance x and the weight of the chain at distance x?

 

[Sheneron]

Ok I figured it out. I wasn't taking into account that fact that the chain was already displacing the spring from rest at the beginning. So I had to alter my limits of integration which yielded the correct result. Thanks.

 

[Sheneron]

By the way, how do you mark a thread as solved?

 

[The_ArtofScience]

I know everyone wants to make this thread an open-and-shut case. If possible, Sheneron, can you give the answer, I'm curious to know.

v^2 = v_0^2 -2aΔx
mv^2 = mv_0^2-2maΔx
2FΔx=mv^2 - mv_0^2
FΔx = 1/2(mv^2 - mv_0^2)

w = 10 ft x 2 lb/ft = 20 lbs
Fxk= 4x
x = 20/4 = 5 ft and a distance of 2 ft makes it 7

1/2(4)(7^2 - 5^2) = 48 lb/ft

 

[Sheneron]

Sure, here is how I solved it.

The force of the spring as a function of x:
f(x) = 4x

The force of the chain as a function of x:
f(x) = 2(10 - x)

The force of the chain is pulling down on the spring so the total force would be:

f(x) = 4x - 2(10 - x)

Now the part the tricksy part to keep in mind is that, the chain is already initially displacing the spring from its natural position. So we need to find out how far the chain pulls the spring down in x direction initially.

So set the two forces equal to each other and solve for x:

4x = 20 - 2x
x = \frac{20}{6}

Since you want it stretched 2 additional feet, you would end up with an integral such as this:

\int_{\frac{20}{6}}^{\frac{32}{6}} 4x - 2(10-x) dx
\int_{\frac{10}{3}}^{\frac{16}{3}} 6x - 20 dx

Integrate that and you get an answer of 12 lb*ft

 

[The_ArtofScience]

Sure, here is how I solved it.

The force of the spring as a function of x:
f(x) = 4x

The force of the chain as a function of x:
f(x) = 2(10 - x)

The force of the chain is pulling down on the spring so the total force would be:

f(x) = 4x - 2(10 - x)

Now the part the tricksy part to keep in mind is that, the chain is already initially displacing the spring from its natural position. So we need to find out how far the chain pulls the spring down in x direction initially.

So set the two forces equal to each other and solve for x:

4x = 20 - 2x
x = \frac{20}{6}

Since you want it stretched 2 additional feet, you would end up with an integral such as this:

\int_{\frac{20}{6}}^{\frac{32}{6}} 4x - 2(10-x) dx
\int_{\frac{10}{3}}^{\frac{16}{3}} 6x - 20 dx

Integrate that and you get an answer of 12 lb*ft

That's a very nice solution, Sheneron. I assumed that k = 4lb/ft without realizing that it was being displaced at rest