# Mechanical Energy and Newton's Laws

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1. Sep 5, 2016

### DumSpiroSpero

1. The problem statement, all variables and given/known data

A block of mass $m$, attached to a rope, is dropped at the point A.
When the block reaches the point B, the tension is $T = 2 \cdot m \cdot g$ and the rope broke at that same point.
If the lenght of the rope is $L = 6cm$, evaluate the height $h$ where the rope broke.

2. Relevant equations
The solution's manual uses mechanical energy and the centripetal force ( $T - m \cdot g \cdot sin\theta = \frac{m \cdot v^{2}}{2}$). Since $h = sin \theta \cdot L$, we find the height $h = 4cm$.

3. The attempt at a solution
I understand and agree with that solution, but I can't figure out why my solution is wrong.
Since $W = m \cdot g$ is perpendicular to the horizontal line, we have a vectorial triangle composed by $\vec{T}, \vec{W}, \vec{F}$ (the green, red and purple vectors in the figure below, respectively), where F is the resultant force of T and W, i.e. $\vec{F} = \vec{T} + \vec{P}$ .
In this triangle, we have $sin \theta = \frac{W}{T} = \frac{W}{2\cdot W} = \frac{1}{2}$ and $sin \theta = \frac{h}{L}$. Therefore, $h = \frac{L}{2} = 3cm$. I couldn't find my mistake.

2. Sep 5, 2016

### TSny

Welcome to PF!

You are assuming the net force, F, on the block is horizontal. If that were true in general, would the block ever have any vertical acceleration?

3. Sep 5, 2016

### DumSpiroSpero

Thank you very much - for the welcome and for the solution. :)

After your comment, I noticed that in my figure the tension and the weight have similar lenghts. I redone it with more precise lenghts (T = 2W) and, indeed, it's impossible to the resultant vector to be horizontal in terms of vectors. (R is the real resultant vector, and F is the wrong one at my first attempt.)

4. Sep 6, 2016

OK, good.