Answer:T=20.62 N
Step-by-step explanation:
Given
[tex]m_1=5.8 kg[/tex]
[tex]m_2=1.7 kg[/tex]
inclination [tex]\theta =37^{\circ}[/tex]
Let T be the tension
From diagram
[tex]5.8 g\sin\theta -T=5.8 a[/tex]
[tex]5.8 g\sin\theta -5.8 a=T[/tex]---1
For [tex]m_2[/tex] block
[tex]T-1.7 g=1.7 a[/tex]
[tex]T=1.7(g+a)[/tex]--2
from 1 & 2 we get
[tex]1.7(g+a)=5.8 g\sin\theta -5.8 a[/tex]
[tex]7.5 a=5.8g\sin \theta -1.7g[/tex]
[tex]7.5 a=1.79 g[/tex]
[tex]a=\frac{1.79 g}{7.5}=0.238 g=2.33 m/s^2[/tex]
[tex]T=1.7(9.8+2.33)=1.7\times 12.132[/tex]
[tex]T=20.62 N[/tex]