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A man pulling sled of his daughter by a massless rope, climbing a snowy hill whose slope is equal to 15 °. Considering that the mass of the sled is $4Kg$, the girl's $26Kg$ and $\mu _c = 0,25$, calculate the work done by the tension of the string after walking with constant speed a distance of $130m \Delta x$

So, I know:

$\varphi=15°; \alpha =30°; m_d=26Kg; m_s=4Kg; \mu _c =0,25; d=130m$

So, the equations will be:

Weight in Y:

$P_y=P.\cos\varphi $

Weight in X:

$P_x=P.\sin\varphi$

Tension in Y:

$T_y=T.\sin\alpha$

Tension in X:

$T_x=T.\cos\alpha$

Normal force:

$P_y-N-T_y=0 \implies N = T_y-P_y$

Newton's 2nd law:

$\sum F = m_t.a \implies T_x-F_r-P_x=m_t.a$

Frictional force:

$F_r=\mu _c.N$

work definition:

$W=m_t.a.d$

Total mass:

$m_t=m_d+m_s \implies m_t =30Kg$

I want to get the acceleration of the object and then replacing $a$ I can get the string tension, so (...)

$a={{T_x-F_r-P_x}\over m_t}$

Now, ${T_x-F_r-P_x} = \sum F \implies {T_x-F_r-P_x} = W =m_t.a.d$

But again, If such replacement, the unknown $a$ appears again.

¿How can I solve this?

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1  
If the walking is at constant velocity, what is the acceleration? –  User58220 Oct 10 '13 at 2:32
    
$$\sum F = 0$$ - No acceleration! –  mcodesmart Oct 10 '13 at 2:40
    
Then the Work is 0! Post as an answer, so I'll give you the accepted answer. Thanks –  Tomi Sebastián Juárez Oct 10 '13 at 2:48
    
No, there is still tension in the spring. You need to find it and multiply the distance moved. –  mcodesmart Oct 10 '13 at 2:53

1 Answer 1

up vote 1 down vote accepted

$\varphi=15°$ ; $m_t=30 kg $; $\mu _c =0,25 $; $ d=130m$;

Newton's 1st law:

$$\sum F_x = 0 \implies T_x-\mu _c m_tg cos \varphi - m_tgsin \varphi= 0$$ $$\sum F_y = 0 \implies T_y-m_tg cos \varphi = 0$$

$$\begin{pmatrix} T_x \\ T_y \end{pmatrix} = \begin{pmatrix} \mu _c m_tg cos \varphi +m_tgsin \varphi \\ m_tg cos \varphi \end{pmatrix} = \begin{pmatrix} 1.23 \\ 0.96 \end{pmatrix} N $$

After you find $T_x$ and $T_y$, you can find the resultant vector which will be in the direction of $\vec d$. The work done then is simply:

$$W=\vec T \bullet \vec d = |T||d|$$

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