Change in Kinetic Energy of a Cart + Pulley System

[Flamefury]

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!

 

[rl.bhat]

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.

 

[Flamefury]

This may be a silly question, but how do I integrate x = 3 to x = 1?

 

[rl.bhat]

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.

 

[Flamefury]

I understand up to subbing in the cosθ into the dW equation. However, I don't understand the process of integration.

 

[rl.bhat]

Write down the expression for dW.

 

[Flamefury]

dW = F(\frac{x}{\sqrt{x^{2}+h^{2}}})dx

 

[rl.bhat]

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

 

[Flamefury]

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.
W=F*(\sqrt{x_{i}^{2}+h^{2}}-\sqrt{x_{f}^{2}+h^{2}})

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.