Question

1. if masses m₁ and m₂ on the inclined friction - less surface are pulled by a force f = 40 n, as shown in the figure, m₁ = 2 kg m₂ = 3 kg a) what is the acceleration of the system b) find the tension in the string connecting the masses. (sin37 = 0.6, cos37 = 0.8)

Explanation:

Step1: Analyze the forces along the incline

According to Newton's second law \(F_{net}=ma\). The net force acting on the system along the incline is \(F - (m_1 + m_2)g\sin37^{\circ}\), and the total mass of the system is \(m_1 + m_2\).

$$F-(m_1 + m_2)g\sin37^{\circ}=(m_1 + m_2)a$$

Step2: Solve for acceleration \(a\)

Substitute \(F = 40N\), \(m_1=2kg\), \(m_2 = 3kg\), \(g = 10m/s^{2}\), \(\sin37^{\circ}=0.6\) into the above - equation:

$$40-(2 + 3)\times10\times0.6=(2 + 3)a$$

$$40-30 = 5a$$

$$a=\frac{10}{5}=2m/s^{2}$$

Step3: Analyze the forces on \(m_1\)

For mass \(m_1\), using Newton's second law \(T - m_1g\sin37^{\circ}=m_1a\)

Step4: Solve for tension \(T\)

Substitute \(m_1 = 2kg\), \(g = 10m/s^{2}\), \(\sin37^{\circ}=0.6\), \(a = 2m/s^{2}\) into the equation:

$$T-2\times10\times0.6=2\times2$$

$$T - 12=4$$

$$T=16N$$

Answer:

a) The acceleration of the system is \(2m/s^{2}\)
b) The tension in the string is \(16N\)