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The problem is as follows:

The diagram from below shows a block being pulled by a wire. The block's mass is $10\,kg$ and it moves horizontally from point $A$ to point $B$ due a constant force labeled $\vec{F}$ whose modulus is $40\,N$. Find the work done by the force $F$. The distance between $AB$ is $3.5 m$.

Sketch of the problem

The alternatives in my book are:

$\begin{array}{ll} 1.&400\,J\\ 2.&300\,J\\ 3.&140\,J\\ 4.&100\,J\\ \end{array}$

Initially I thought that the work can be found using this formula:

$W=F \cdot d$

Since they mention $F= 40\,N$:

$W=F\cos 37^{\circ}\cdot 3.5=\left(40\right)\left(\frac{4}{5}\right)\left(3.5\right)$

$W=112\,J$

However this doesn't seem right as I believe the work done by pulling the wire is measured by the distance which is traveled by the wire and not by the block.

It is kind of a strange setting as I cannot imagine a block which stays in the ground as is being pulled as it is described.

My instinct tells me that it has something to do with the horizontal distance in the sense that the distance which will be doing the force is the difference between the hypotenuse of the triangle from A to the pulley minus B to the pulley. But these distances aren't exactly given.

By continuing my attempt I spotted these relationships in the triangles as shown in the diagram from below and I could made these equations:

Sketch of the problem

The distance which will be doing work will be given by the difference between the big hypotenuse minus the smaller hypotenuse, in the sense of $AP-BP=d$.

Using the trigonometric identities then I reached to:

$d=\frac{3.5+h\cos 53^{\circ}}{\cos 37^{\circ}}-h$

But for this is required $h$.

To do so. I thought to use:

$\tan 37^{\circ}=\frac{h\sin 53^{\circ}}{3.5+h\cos 53^{\circ}}$

Therefore:

$\frac{3}{4}=\frac{\frac{4h}{5}}{\frac{35}{10}+\frac{3h}{5}}$

Then:

$3\left(\frac{35}{10}+\frac{3h}{5} \right )=\frac{16h}{5}$

$3\left(35+6h\right)=32h$

$105+18h=32h$

$h=7.5$

Therefore with this information the distance can be computed as follows:

$d=\frac{3.5+7.5\cos 53^{\circ}}{\cos 37^{\circ}}-7.5$

$d=\frac{\frac{35}{10}+\frac{75}{10}\frac{3}{5}}{\frac{4}{5}}-\frac{75}{10}$

$d=\frac{\frac{35}{10}+\frac{15}{2}\frac{3}{5}}{\frac{4}{5}}-\frac{75}{10}$

$d= \frac{\frac{80}{10}}{\frac{4}{5}}-\frac{75}{10}$

$d= \frac{400}{40}-\frac{75}{10}=10-7.5=2.5$

Therefore that would be the distance required to calculate the work done by pulling the wire.

By pluggin this number with that of the given force then the work is:

$W=F \cdot d = 40 \cdot 2.5 = 100\,J.$

So this would comply with the fifth option. But does it exist an easier method to this?. I'm still confused at why was I given the weight of that body?. I did not used this number to obtain this answer. Or could it be that i'm overlooking something. Can someone offer some help here?.

This is the part where I'm stuck. Can somebody help me with this please?.

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  • $\begingroup$ The question to ask is: How much does the end point of the rope (where the letter $F$ is written) move? The answer: It's equal to the difference of the lengths of the diagonal line, difference between the final length and the initial length. You can use trigonometry to solve for that. Obviously, the angles $37^{\circ}$ and $53^{\circ}$ are important in this computation. So if $P$ is the point in the pulley where the rope becomes parallel to the ground, the two distances are $AP$ and $BP$. So what is $|AP|-|BP|$? $\endgroup$
    – Matti P.
    Nov 26 '19 at 6:33
  • $\begingroup$ @MattiP. It is exactly this part where I'm stuck at. How exactly can I find that Can you help me?. $\endgroup$ Nov 26 '19 at 11:12
  • $\begingroup$ So you have a triangle. One angle is $37^{\circ}$ and the one next to it is $180-53=127^{\circ}$. The length of the side between these two is $3.5~\text{m}$. The angle that is opposite to this side is $180-37-127 = 16^{\circ}$. Next, you can use the Law of Cosines: en.wikipedia.org/wiki/Law_of_sines $\endgroup$
    – Matti P.
    Nov 26 '19 at 11:32
  • $\begingroup$ @MattiP. But to do so I need the value of one of those sides. I mean the hypotenuse which would be $BP$ or $BA$ and these are both unknown. How exactly the Law of Cosines would help me here?. $\endgroup$ Nov 26 '19 at 13:36
  • $\begingroup$ @MattiP. Perhaps could you add your solution so I can compare it with my own?. I think I solved it as I arrived to $100\,J$ but I'm not sure if that would be the answer. $\endgroup$ Nov 26 '19 at 13:37
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We can use the law of sines. The angle at the peak of the triangle is $16^\circ$ so we have $$\frac {3.5}{\sin 16^\circ}=\frac h{\sin 37^\circ}\\ h \approx 7.6417$$ Similarly, if $k$ is the left hand edge of the triangle $$\frac {3.5}{\sin 16^\circ}=\frac k{\sin 127^\circ}\\ k \approx 10.1409$$ The difference of these, to one decimal, is $2.5$ and the work is $40\ N \cdot 2.5\ m=100J$ You don't need the mass of the block. If the block is lighter, you will pull faster so its final velocity is higher making the kinetic energy $100\ J$.

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Use the relationships, $$d_2-d_1 = h\csc37 - h\csc53,\>\>\>\>\> 3.5= h\cot 37 - h\cot 53$$

Eliminate $h$ to get,

$$d_2-d_1 = 3.5\cdot\frac{\csc37 - \csc53}{\cot 37 - \cot 53}=3.5\cdot \frac{\frac53-\frac54}{\frac43-\frac34}=2.5$$

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Hint.

Calling $\phi_1 = 37,\ \phi_2 = 53,\ l_0 = 3.5$ we have

$$ \cases{ l_1\sin\phi_1=l_2\sin\phi_2\\ l_1\cos\phi_1-l_2\cos\phi_2 = l_0\\ d = l_1-l_2\\ W = F d } $$

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