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I think I am making this Work-Energy Problem to Hard

  • Thread starter curly_ebhc
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Homework Statement



Figure 7-40 shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.6 m, so the cart slides from x1 = 4.0 m to x2 = 1.0 m. During the move, the tension in the cord is a constant 27.0 N. What is the change in the kinetic energy of the cart during the move?

Picture Attached
Answer: 65.376061721000Units J


Homework Equations


Work energy thorem Delta K=W= Fd cos (theta)


The Attempt at a Solution



The first thing I tried with a friend was simply fdcos (theta) but then we realized that the angle is constantly changing. So we integrated Fd cos(theta) d(theta) from theta(at x1) to theta(at x2) but we got the wrong answer. We were stumped and so we asked another friend who reminded us that when you integrate d(theta) that you are integrating over an arc length rather than a linear distance. So we figured that you can turn the distance into a function of theta and it is no longer a constant. Or integrate over distance and make theta a function of distance. This seems like it would work but it gives a really gnarly integral with a cos and a tangent. That brings up red flags that it should be easier than that for an intro course at a JC where we only sometimes do basic integrals.

Thanks

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
Delphi51
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It IS a hard problem. I hope you know calculus.
Write an expression for the bit of work dW done when the block is pulled a small distance dx. Integrate it from x1 to x2. Remember that only the horizontal component of the tension does work on the block.
 
  • #3
collinsmark
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The first thing I tried with a friend was simply fdcos (theta) but then we realized that the angle is constantly changing. So we integrated Fd cos(theta) d(theta) from theta(at x1) to theta(at x2) but we got the wrong answer.
That approach should give you the right answer. So maybe you didn't integrate correctly. Show your steps.
We were stumped and so we asked another friend who reminded us that when you integrate d(theta) that you are integrating over an arc length rather than a linear distance. So we figured that you can turn the distance into a function of theta and it is no longer a constant. Or integrate over distance and make theta a function of distance. This seems like it would work but it gives a really gnarly integral with a cos and a tangent. That brings up red flags that it should be easier than that for an intro course at a JC where we only sometimes do basic integrals.
By the way, it is possible to solve this problem without any calculus (well, almost no calculus).
  • What is the definition of work?
  • What is the difference in cord length between the two events?
There is work being applied to the cord (regardless of the system's configuration) over some length of cord, by the person/mechanism pulling on the rope. Since the system's potential energy doesn't change, and since there is no friction (and since the cord and pulley can be assumed massless), where does that energy go? :wink:
 
  • #4
Delphi51
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I don't think the non-calculus method will work, collinsmark. The difficulty is that the vertical component of the tension does no work; it just tries to lift the block. The horizontal component doing the work varies with x.
An average value of T*cos(θ) would be better, though. Perhaps that is the expected approach because the integral is not very basic. But you can easily look it up in a table or at integrals.com.

Show us your work, Curly, and we'll make sure you get it right!
 
  • #5
collinsmark
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I don't think the non-calculus method will work, collinsmark. The difficulty is that the vertical component of the tension does no work; it just tries to lift the block. The horizontal component doing the work varies with x.
An average value of T*cos(θ) would be better, though. Perhaps that is the expected approach because the integral is not very basic. But you can easily look it up in a table or at integrals.com.

Show us your work, Curly, and we'll make sure you get it right!
Oh, but it does solve the problem. :smile: The vertical component of the tension does no work, I agree, but it doesn't cause the block to move either. (And thus it doesn't cause any rope to move past the pulley.)

Trust me. I stand by what I say on this: the problem can be solved without calculus. (Edit: well, maybe really trivial calculus such as ∫dx, but that's it though.)
 
Last edited:
  • #6
Delphi51
Homework Helper
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Absolutely right, Collinsmark!
I worked it out both ways and got the same answer.
I guess energy is conserved after all!!

It is different from the usual problem of pulling a sled with a rope at a vertical angle where you must use only the horizontal component to find work done.
 

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