Change in Kinetic Energy of a Cart + Pulley System

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Homework Help Overview

The problem involves a cart sliding along a frictionless rail connected to a pulley system, where the change in kinetic energy is to be determined as the cart moves from one position to another while a constant tension is applied to the cord. The context includes the calculation of work done using the angle of the cord and the displacement of the cart.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of work done, questioning the assumption of a constant angle θ during the cart's movement. There are attempts to express cosθ in terms of the cart's position and height, and to set up the integration needed to find the work done over the specified limits.

Discussion Status

Some participants have provided guidance on how to express the work done in terms of integration, while others express uncertainty about the integration process itself. There is an acknowledgment of different approaches to the problem, with one participant suggesting a method based on the change in hypotenuse as displacement.

Contextual Notes

Participants are navigating the complexities of integrating a variable angle in the work equation and are discussing the implications of the pulley system on the calculations. There is a mix of understanding regarding the integration process and the assumptions made in the initial calculations.

Flamefury
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Homework Statement


http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_42.gif
Figure 7-42 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 = 1.20 m, so the cart slides from x1 = 3.00 m to x2 = 1.00 m. During the move, the tension in the cord is a constant 25.0 N. What is the change in the kinetic energy of the cart during the move?

Homework Equations


W = F*d cosθ

The Attempt at a Solution


Find θ first.
θ = tan-1(h / Δx)
= tan-1(1.2 / (3-1))
= tan-1(0.6)

Sub into work equation.
W = F*d*cosθ
= (25) * (2) * cos (tan-1(0.6))
= 42.87464629 J

The answer should be 41.7 J. Help is much appreciated!
 
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Hi Flamefury, welcome to PF.
You have calculated work done assuming that θ remains constant. But it is not so.
Expess
dW = Fcosθ*dx
write down cosθ in terms of h and x. Find the integration from x = 3 to x = 1.
 
This may be a silly question, but how do I integrate x = 3 to x = 1?
 
First of write down cosθ in terms of h and x.
Substitute it in dW. Find the integration between the limits x = 3 m to x = 1 m.
 
I understand up to subbing in the cosθ into the dW equation. However, I don't understand the process of integration.
 
Write down the expression for dW.
 
[tex]dW = F(\frac{x}{\sqrt{x^{2}+h^{2}}})dx[/tex]
 
Do you know how to find integration?
It can be done by substitution method.
Put t = x^2 + h^2
dt = 2x*dx
x*dx = dt/2
 
I still don't understand the integral, but I was able to get the answer.

I wrote the work equation using the change in hypotenuse as the displacement. This works because of the pulley at the top, and the fact that the string can't be created or destroyed.
[tex]W=F*(\sqrt{x_{i}^{2}+h^{2}}-\sqrt{x_{f}^{2}+h^{2}})[/tex]

So I just used this to figure out the work done.

But I thank you for your help, regardless! I appreciate how long you put up with me for.
 

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