How do I find the work done by a force pulling a block from a pulley in a ceiling?

Question

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 $3.5 m$.

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:

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?.

Answer 1

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 $100\ J$.

Answer 2

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

Eliminate $h$ to get,

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

Answer 3

Hint.

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

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