The problem is as follows:

The alternatives given on my book are as follows:

$\begin{array}{ll}

1.&\left\langle 30,-30\right\rangle\,N\\

2.&\left\langle 45,25\right\rangle\,N\\

3.&\left\langle 70,40\right\rangle\,N\\

4.&\left\langle 40,-40\right\rangle\,N\\

5.&\left\langle 15,10\right\rangle\,N\\

\end{array}$

What I thought to do in this problem was to use vectors as follows:

This summarized in the figure from below:

Using $\sin\phi=\frac{4}{5}$ then $\cos\phi=\frac{3}{5}$

Acoording to Newton's second law the sum of forces must be mass times acceleration as:

$\sum_{i=1}^{n}F=ma$

$F_{1}=\left \langle 0,F_{1} \right \rangle$

$F_{2}=\left \langle x,y \right \rangle$

$a=\left \langle a\cos\phi,a\sin\phi \right \rangle$

Therefore:

$\sum_{i=1}^{n}F=m\left(\left \langle a\cos\phi,a\sin\phi \right \rangle\right)$

$\left \langle 0,F_{1} \right \rangle+\left \langle x,y \right \rangle=m\left \langle a\cos\phi,a\sin\phi \right \rangle$

For $x$:

$x=m a\cos\phi = 5 \times 10 \left(\frac{3}{5}\right)=30$

For $y$:

$F_{1}+y=m a\sin\phi$

$y= m a\sin\phi - F_{1} = 5 \times 10 \left( \frac{4}{5}\right)-70$

$y= 40-70 = -30$

Therefore:

$F_{2}=\left \langle x,y \right \rangle = \left\langle 30,-30 \right \rangle$

This looks acceptable and it checks with the alternative $1$ but I am not sure if what I did is right or not. Can someone help me to verify this?.

A body of $5\,kg$ of mass is at rest when $x=0$. suddenly, two forces $\vec{F_{1}}$ and $\vec{F_{2}}$ act on the body applying on it an acceleration of $10\frac{m}{s^{2}}$ in magnitude. It is known that $F_{1}=70\,N$. Find the force $F_{2}$ in $N$. You may consider that $\sin\phi=\frac{4}{5}$.

The alternatives given on my book are as follows:

$\begin{array}{ll}

1.&\left\langle 30,-30\right\rangle\,N\\

2.&\left\langle 45,25\right\rangle\,N\\

3.&\left\langle 70,40\right\rangle\,N\\

4.&\left\langle 40,-40\right\rangle\,N\\

5.&\left\langle 15,10\right\rangle\,N\\

\end{array}$

What I thought to do in this problem was to use vectors as follows:

This summarized in the figure from below:

Using $\sin\phi=\frac{4}{5}$ then $\cos\phi=\frac{3}{5}$

Acoording to Newton's second law the sum of forces must be mass times acceleration as:

$\sum_{i=1}^{n}F=ma$

$F_{1}=\left \langle 0,F_{1} \right \rangle$

$F_{2}=\left \langle x,y \right \rangle$

$a=\left \langle a\cos\phi,a\sin\phi \right \rangle$

Therefore:

$\sum_{i=1}^{n}F=m\left(\left \langle a\cos\phi,a\sin\phi \right \rangle\right)$

$\left \langle 0,F_{1} \right \rangle+\left \langle x,y \right \rangle=m\left \langle a\cos\phi,a\sin\phi \right \rangle$

For $x$:

$x=m a\cos\phi = 5 \times 10 \left(\frac{3}{5}\right)=30$

For $y$:

$F_{1}+y=m a\sin\phi$

$y= m a\sin\phi - F_{1} = 5 \times 10 \left( \frac{4}{5}\right)-70$

$y= 40-70 = -30$

Therefore:

$F_{2}=\left \langle x,y \right \rangle = \left\langle 30,-30 \right \rangle$

This looks acceptable and it checks with the alternative $1$ but I am not sure if what I did is right or not. Can someone help me to verify this?.

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