Given:
Mass, m1 = 10.0 kg
Mass, m2 = 3.60 kg
θ = 37.0 degrees
Let's solve for the following:
(a) Magnitude of the acceleration of 3.60 kg box.
From Newton's second law, we have:
Fnet = ma
m₁g - T = m₁a................equation 1
Applying Newton's second law along an inclined plane, we have:
T - m₂gsinθ = m₂a.............equation 2
Add both equation 1 and 2:
m₁g - T + T - m₂gsinθ = m₁a + m₂a
m₁g - m₂gsinθ = a(m₁ + m₂)
To find the magnitude of aceleration for m2, rewrite the equation for a:
[tex]\begin{gathered} a=\frac{m_{1}g-m_{2}gsin\theta}{m_{1}+m_{2}} \\ \\ a=\frac{g(m_1-m_2sin\theta)}{m_1+m_2} \end{gathered}[/tex]
Where:
g is acceleration due to gravity = 9.8 m/s²
[tex]\begin{gathered} a=\frac{9.8(10-3.6\sin 37)}{10+3.6} \\ \\ a=5.64m/s^2 \end{gathered}[/tex]
The manitude of acceleration of the 3.60 kg box is 5.64 m/s².
(b). What is he tension in the cable?
To find the tension in the cable, apply the formula:
[tex]\begin{gathered} m_1g-T=m_1a \\ \\ T=m_1g-m_1a \\ \\ T=m_1\mleft(g-a\mright) \\ \\ T=10.0(9.8-5.64) \\ \\ T=41.6\text{ N} \end{gathered}[/tex]
The tension in the cable is 41.6 N
ANSWER:
(a) 5.64 m/s²
(b) 41.6 N
