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

Jun 2017
229
6
Lima, Peru
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.

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

skeeter

Math Team
Jul 2011
3,135
1,698
Texas
note that the horizontal component of force that is performing work is $F\cos{\theta}$, where $\theta$ is variable. $\theta$ increasing $\implies \cos{\theta}$ is decreasing.

The initial component is 32 N and the final component is 24 N. Taking the average, 28 N, and using that value to determine the work over the displacement of 3.5 M yields an approximation of 98 J.

To get the exact value requires an integral ...

$\displaystyle W = \int_0^{3.5} 40 \cdot \dfrac{8-x}{\sqrt{(8-x)^2+6^2}} \, dx = 100 \text{ N}$

details are in the attached diagram.
 

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Jun 2017
229
6
Lima, Peru
By continuing my attempt I spotted these relationships in the triangles in the new diagram 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?. :help:
 
Jun 2017
229
6
Lima, Peru
note that the horizontal component of force that is performing work is $F\cos{\theta}$, where $\theta$ is variable. $\theta$ increasing $\implies \cos{\theta}$ is decreasing.

The initial component is 32 N and the final component is 24 N. Taking the average, 28 N, and using that value to determine the work over the displacement of 3.5 M yields an approximation of 98 J.

To get the exact value requires an integral ...

$\displaystyle W = \int_0^{3.5} 40 \cdot \dfrac{8-x}{\sqrt{(8-x)^2+6^2}} \, dx = 100 \text{ N}$

details are in the attached diagram.
Gee.. I never thought that I could use integrals as complex as this.

Btw I've added a calculus free solution which does arrive to the same answer that you got.

But yours looks more "fancy". I got the part of how you calculated the sides in the triangle. But i'm stuck of where does $\left(8-x\right)$ comes from?.

The integral you posed can be done by hand?. I have forgotten the integration methods, but it looks very likely to be by trigonometric substitution?.

Wolfram alpha says is:

see here.

$\frac{2}{3}\sqrt{44-x}\left(x+64\right)+C$

I don't know exactly how it got to this solution. How did you obtained it?. :eek:
 

skeeter

Math Team
Jul 2011
3,135
1,698
Texas
I used a calculator ... see attached.

The integral is doable by hand , might require a couple of substitutions.

Doing this on the fly ...

\(\displaystyle \int_0^{3.5} \dfrac{40(8-x)}{\sqrt{(8-x)^2+36}} \, dx\)

$u = 8-x \implies du = -dx$

\(\displaystyle \int_{4.5}^8 \dfrac{40u}{\sqrt{u^2+36}} \, du\)

one more substitution ...

$v = u^2+36 \implies dv = 2u \, du$

\(\displaystyle 20 \int_{56.25}^{100} \dfrac{1}{\sqrt{v}} \, dv\)

$40 \bigg[\sqrt{v} \bigg]_{56.25}^{100} = 40(10 - 7.5) = 100$
 

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skeeter

Math Team
Jul 2011
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The law of sines could also be used to determine the length of wire pulled through the pulley ...
 

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Jun 2017
229
6
Lima, Peru
I used a calculator ... see attached.

The integral is doable by hand , might require a couple of substitutions.

Doing this on the fly ...

\(\displaystyle \int_0^{3.5} \dfrac{40(8-x)}{\sqrt{(8-x)^2+36}} \, dx\)

$u = 8-x \implies du = -dx$

\(\displaystyle \int_{4.5}^8 \dfrac{40u}{\sqrt{u^2+36}} \, du\)

one more substitution ...

$v = u^2+36 \implies dv = 2u \, du$

\(\displaystyle 20 \int_{56.25}^{100} \dfrac{1}{\sqrt{v}} \, dv\)

$40 \bigg[\sqrt{v} \bigg]_{56.25}^{100} = 40(10 - 7.5) = 100$
Thanks for that. As I mentioned I forgotten this, I was about to go to my old trusty Anton's Calculus for that. :)

But you forgot to mention how did you obtained $8-x$. I'm guessing is due to the $3-4-5$ triangle since one side is $6$ the other is $8$ and the displacement is given by $x$ then the new distance is $8-x$.

I'm still confused. I could naively said okay $\theta = 53^{\circ}$ but why this isn't?. Isn't given that the block moves from $A$ to $B$?. Can you explain these parts to me please? (Note: I'm using boldface to this as since the forum has been updated I haven't found an icon to replace the older one which isn't available.).
 

skeeter

Math Team
Jul 2011
3,135
1,698
Texas
I'm guessing is due to the 3−4−5 triangle since one side is 6 the other is 8 and the displacement is given by x then the new distance is 8−x.
yes ... see the blue labeled triangle at the far right of the attached diagram from post #2

the block slides a distance x from point A, forming a right triangle with base $(8-x)$, height $6$, and hypotenuse $\sqrt{(8-x)^2+6^2}$. The acute angle, $\theta$, formed by the wire with the horizontal has $\cos{\theta} = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{8-x}{\sqrt{(8-x)^2+6^2}}$

Therefore, an increment of work done, $\displaystyle dW = 40 \cdot \dfrac{8-x}{\sqrt{(8-x)^2+6^2}} \, dx \implies W = 40 \int_0^{3.5} \dfrac{8-x}{\sqrt{(8-x)^2+6^2}} \, dx$