Work problem on winding up a cable

[Dan350]

Find the work done in winding up a 300ft cable that weighs 2.00 lb /ft




W=Fd
W=$$\int F dx$$






My attempt

w=F*D
so W= (300ft)(2.00lb/ft)
W=600ft.lb

and


$$\int_0^{300}\!\ 2xdx$$


W=90,000 ft.lb

I'm taking Calc 2 and this is applied Physics problems.

Which of these answers is correct?
And Why?

Thanks in advance
Cheers!

 

[sonnyfab]

The equation W=Fd only applied when F is constant. Because the force of gravity will decrease on the cable as you continue to pull it up (it will weight less the more your pull it up), you cannot use this equation. You do need to use an integral.

However, the force is not 2x. The force is the weight, which is the (linear density)*(acceleration due to gravity)*dL

 

[Dick]

Find the work done in winding up a 300ft cable that weighs 2.00 lb /ft




W=Fd
W=$$\int F dx$$






My attempt

w=F*D
so W= (300ft)(2.00lb/ft)
W=600ft.lb

and


$$\int_0^{300}\!\ 2xdx$$


W=90,000 ft.lb

I'm taking Calc 2 and this is applied Physics problems.

Which of these answers is correct?
And Why?

Thanks in advance
Cheers!

The second one. The force isn't constant as you reel in the chain. It gets lighter as more is reeled in. And for another thing (300ft)(2.00lb/ft) doesn't give you units of ft*lb.

 

[Dan350]

The equation W=Fd only applied when F is constant. Because the force of gravity will decrease on the cable as you continue to pull it up (it will weight less the more your pull it up), you cannot use this equation. You do need to use an integral.

However, the force is not 2x. The force is the weight, which is the (linear density)*(acceleration due to gravity)*dL

So do I have to use the hooke's law to get the product of the integral?
F= Kx
letting F=2lb/ft and solve for k?
Or I would use 9.8 as an acceleration due gravity?
Thank again

 

[Dick]

So do I have to use the hooke's law to get the product of the integral?
F= Kx
letting F=2lb/ft and solve for k?
Or I would use 9.8 as an acceleration due gravity?
Thank again

You integrate force over the distance. The force depends on the amount of cable you have to yet to reel up. Hooke's law has nothing to do with this. And you don't need 9.8 m/s^2. That's a metric unit. You are working in lbs and ft. Your second solution with the integral of 2x is correct. Try and figure out why. F isn't 2lb/ft. F is 2lb/ft times number of feet.

 

[sonnyfab]

Dan350,
You need to know which force is doing the work (of in this case, which force work is being done against) to figure out what F is. Hooke's Law applies to "restoring forces" which pull objects towards an equilibrium position. The key point about Hooke's Law forces is that whether you are to the felt or right of the equilibrium point, or above or below it, the force points towards that point. Springs pull you back or push you forward, depending on which direction the equilibbrium point of the spring is, and as a result springs obey Hooke's Law.
Your problem has something pulling up a chain. So to figure out which force you are working against, you need to see which force is pulling in the other direction. You pull up and ______ pulls down. I think we can agree that _____ is gravity here.
So F = weight (measured in lb). However, you do not know weight. Instead, you are given (linear weight - or weight per unit length) which is weight/L, the weight divided by the length of the chain. F is not constant in your problem, because the L gets shorter as you continue to pull. The fact that the force is not constant tells you that you must use an integral (it is true that you must also integrate when you use Hookes Law, because F=kx, and so it is not constant but depends on x).

Mass = Linear density * Length, so you have F = (linear weight)*L. It is L that changes as the chain is pulled up, so you are integrating over dL. The smallest L (lower limit) is 0ft and the largest (upper limit) is 300ft.

$\int_0^{300}\!\ (Linear Weight) * dL$

Linear Weight is 2 lb/ft. There is no x inside the integral here.

 

[sonnyfab]

Sorry, I wasn't paying attention to the units. You do need an x inside the integral. The work is the integral of Fdx. You need to multiply mass/length*length to get F, so 2x*dx

Dr Peter Vaughan
BASIS Peoria Physics

 

[Dick]

Sorry, I wasn't paying attention to the units. You do need an x inside the integral. The work is the integral of Fdx. You need to multiply mass/length*length to get F, so 2x*dx

Dr Peter Vaughan
BASIS Peoria Physics

Indeed it is. Though in these units it's weight/length * length=weight=force.