Question about the Work-Kinetic Energy Theorem - Pulling a cart

[titansarus]

Homework Statement

Question from Fundamentals of Physics (Halliday, Resnick, Walker)
This figure below 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

Homework Equations

$\int dw = \int F . dr$

The Attempt at a Solution

At first we get that $x/1.2 = cot \theta$ so $dx = -1.2 (1 + cot ^ 2 \theta) d\theta$

Now I don't know whether to integrate the FIRST Or SECOND: (angles from $x_1=3$ to $x_2=1$ change from$\theta_1 = 0.380506377$ to $\theta_2 = 0.87605805059$ radian)

FIRST: $W= \int dw = \int _3^1 25 . dx = \int _{0.380}^{0.876} 25 * 1.2 (1 + cot ^ 2 \theta) d\theta = 41.67 J$

or using $$F = 25 cos \theta$$ and
SECOND: $W= \int dw = \int _3^1 25 cos\theta . dx = \int _{0.380}^{0.876} 25 * cos \theta * 1.2 (1 + cot ^ 2 \theta) d\theta = 34.77 J$

The book itself solved it with another solution using pythagoras theorem and got the first answer but I want to solve it using angles and don't know why the first one gives the answer of the book. I think second one makes more sense.

 

[kuruman]

The setup for the first integral is incorrect. The correct expression is $dW=\vec{F}\cdot d\vec{s}=F~\cos \theta ~dx$.
The setup for the second integral is correct for the reason above.
Having said that, note that $dx=-h \csc^2 (\theta)d\theta=-h\frac{d\theta}{\sin^2(\theta)}$. It makes the integral calculation easier.

 

[Delta2]

You have some calculation errors in both cases and also the first integral formula is wrong. The second integral is the correct one, and its correct evaluation is 41.7262 according to wolfram
http://www.wolframalpha.com/input/?i=integral+25*1.2*cosx*(1+cot^2x)dx+from+0.380506377+to+0.87605805059

EDIT: I used x entering the integral expression at wolfram but I meant $\theta$, its the same thing for wolfram purposes.

 

[titansarus]

The setup for the first integral is incorrect. The correct expression is $dW=\vec{F}\cdot d\vec{s}=F~\cos \theta ~dx$.
The setup for the second integral is correct for the reason above.
Having said that, note that $dx=-h \csc^2 (\theta)d\theta=-h\frac{d\theta}{\sin^2(\theta)}$. It makes the integral calculation easier.

You have some calculation errors in both cases and also the first integral formula is wrong. The second integral is the correct one, and its correct evaluation is 41.7262 according to wolfram
http://www.wolframalpha.com/input/?i=integral+25*1.2*cosx*(1+cot^2x)dx+from+0.380506377+to+0.87605805059

EDIT: I used x entering the integral expression at wolfram but I meant $\theta$, its the same thing for wolfram purposes.

Yes It seems my answer got wrong because I forgot to multiply 34.77 by 1.2 in calculations. And because the answer of the first one get numerically close to the book answer and the actual answer of the book was approximately 42, I thought the second one is Incorrect.

Thanks.

 

[kuruman]

Yes It seems my answer got wrong because I forgot to multiply 34.77 by 1.2 in calculations. And because the answer of the first one get numerically close to the book answer and the actual answer of the book was approximately 42, I thought the second one is Incorrect.

You could have saved yourself some trouble if you postponed the numerical substitution until the very end. It's a good habit. There is no need to find any angles. See solution below
$$W=\int_{x_2}^{x_1}F \cos \theta~dx=-Fh\int_{\theta_2}^{\theta_1} \cos \theta~\frac{d\theta}{\sin^2\theta}$$Let $u=\sin \theta$. Then $\cos \theta~d\theta=du$. So$$W=-Fh\int_{u_2}^{u_1} \frac{du}{u^2}=+Fh \left(\frac{1}{u_1}-\frac{1}{u_2}\right)=Fh \left(\frac{1}{\sin \theta_1}-\frac{1}{\sin \theta_2}\right).$$From the diagram, $\sin \theta_i=\frac{opp}{hyp}=\frac{h}{\sqrt{x_i^2+h^2}}$. Substitute in the equation above to get,$$W=F \left(\sqrt{x_1^2+h^2}-\sqrt{x_2^2+h^2}\right)=25(N)\times \left(\sqrt{3^2+1.2^2}(m)-\sqrt{1^2+1.2^2}(m)\right)=41.73~J$$