Question

I need help with question 7, I just need the answer you don’t have to explain.

Answer 1

`\begin{gathered} a=5.703\text{ }(m)/(s^2) \\ T=3.891\text{N} \end{gathered}`

Step-by-step explanation

Step 1

free body diagram

so, for m1

m1= 10 kg

so,

`\begin{gathered} \sum ^(forces)_{\text{ y}}=T_1-mg=ma \\ T_1=m_1(a+g)\Rightarrow equation(1) \\ T_1=10(a+g) \end{gathered}`

and for m2

`\begin{gathered} \sum ^(forces)_{\text{ x}}=m_2g\sin (37)-T_1=m_2a \\ \text{solve for T}_1 \\ T_1=m_2g\sin (37)-m_2a \\ \text{replace} \\ T_1=(3.6\text{ kg)(9.8 }(m)/(s^2))\sin 37-3.6a \\ T_1=21.23\text{ -}3.6a\Rightarrow eq(2) \end{gathered}`

Step 2

now, replace in equaiotn (1) and solve for a

`\begin{gathered} T_1=m_1(a+g)\Rightarrow equation(1) \\ 21.23\text{ -}3.6a=10(a+g) \\ 21.23\text{ -}3.6a=10a+10g \\ -7.2a=10a-98.1 \\ -17.2a=-98.1 \\ a=-(98.1)/(-17.2) \\ a=5.703\text{ }(m)/(s^2) \end{gathered}`

finally, replace in equation (2) to find Tension

`\begin{gathered} T_1=21.23\text{ -}3.6a \\ T_1=21.23\text{ -}3.6(5.703) \\ T_2=3.891\text{N} \end{gathered}`

I hope this helps you