The problem is as follows:

The figure from below shows a block which is pulled from point $A$ to point $B$ by a force $F=50\,N$ and a constant direction. Find the work in Joules that is made between points $A$ and $B$. (Hint: You may use the triangle $7-24-25$ for $16^{\circ}-74^{\circ}-90^{\circ}$)

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&-310\,J\\ 2.&-250\,J\\ 3.&+310\,J\\ 4.&+250\,J\\ 5.&+280\,J\\ \end{array}$

Since the angle they use is $8^{\circ}$.

I could use the identity for half angle to obtain the relationships in the given triangle.

$\sin 8^{\circ}=\sqrt{\frac{1-\cos 16^{\circ}}{2}}=\sqrt{\frac{1-\frac{24}{25}}{2}}$

$\sin 8^{\circ}=\sqrt{\frac{\frac{1}{25}}{2}}=\frac{\sqrt{2}}{10}$

$\cos 8^{\circ}=\sqrt{\frac{\frac{49}{25}}{2}}=\frac{7\sqrt{2}}{10}$

But other than that I'm still stuck.

I can tell the distance between $A$ and $B$ as:

$AB=7\sec 8^{\circ}=\frac{7}{\cos 8^{\circ}}=\frac{7}{\frac{7\sqrt{2}}{10}}=\frac{10}{\sqrt{2}}=5\sqrt{2}$

However that's how far I went with this problem.

I'm stuck as I don't know how to use the information provided of the force with the angle given.

My intuition tells me that I could naively say okay:

$W= F\times d = 50\cos 37^{\circ} \times 5 \sqrt{2}$

But I'm certain that this will not be the answer and neither appears in the alternatives. Can somebody help me here please?.


1 Answer 1


angle $\theta +53^o + 8^o = 90^o $ by the Figure $ \theta = 29^o $

So $ W = F \times d = 50 \cos 29^o \times 5 \sqrt{2} = 309.22476293882954 $


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