91Ó°ÊÓ

The gravitational force acting on each block can be calculated using the formula \(F_g = mg\), where \(m\) is the mass of the block and \(g\) is the acceleration due to gravity (approximately \(9.81\,\mathrm{m/s^2}\)). For block 1, \(F_{g1} = m_1g = 1.23\,\mathrm{kg} \times 9.81\,\mathrm{m/s^2} = 12.08\,\mathrm{N}\). For block 2, \(F_{g2} = m_2g = 2.46\,\mathrm{kg} \times 9.81\,\mathrm{m/s^2} = 24.15\,\mathrm{N}\).

We will divide these gravitational forces into two components - one parallel to the inclined plane (\(F_{g \parallel}\)) and one perpendicular to the inclined plane (\(F_{g \perp}\)). We can do this using the angle of inclination and basic trigonometry. Parallel component: \(F_{g \parallel} = F_g \sin \theta\) Perpendicular component: \(F_{g \perp} = F_g \cos \theta\) For block 1: \(F_{g1 \parallel} = 12.08\,\mathrm{N} \times \sin(40^{\circ}) = 7.76\,\mathrm{N}\) \(F_{g1 \perp} = 12.08\,\mathrm{N} \times \cos(40^{\circ}) = 9.26\,\mathrm{N}\) For block 2: \(F_{g2 \parallel} = 24.15\,\mathrm{N} \times \sin(40^{\circ}) = 15.5\,\mathrm{N}\) \(F_{g2 \perp} = 24.15\,\mathrm{N} \times \cos(40^{\circ}) = 18.5\,\mathrm{N}\)

The frictional force can be calculated using the formula \(F_f = \mu F_{g \perp}\), where \(\mu\) is the coefficient of kinetic friction. For block 1, \(F_{f1} = \mu_1 F_{g1 \perp} = 0.23 \times 9.26\,\mathrm{N} = 2.13\,\mathrm{N}\) For block 2, \(F_{f2} = \mu_2 F_{g2 \perp} = 0.35 \times 18.5\,\mathrm{N} = 6.48\,\mathrm{N}\)

To find the net force acting on the system, we subtract the frictional forces from the parallel gravitational forces for both blocks: \(F_{net} = (F_{g1 \parallel} - F_{f1}) + (F_{g2 \parallel} - F_{f2}) = (7.76 - 2.13)\,\mathrm{N} + (15.5 - 6.48)\,\mathrm{N} = 14.65\,\mathrm{N}\)

According to Newton's second law, \(F = ma\). Since we have the net force acting on the system (\(F_{net}\)) and the total mass of the system (\(m_{total} = m_1 + m_2\)), we can calculate the acceleration of the blocks: \(14.65\,\mathrm{N} = (1.23\,\mathrm{kg} + 2.46\,\mathrm{kg}) \times a\) Solving for \(a\): \(a = \frac{14.65\,\mathrm{N}}{(1.23+2.46)\,\mathrm{kg}} = \frac{14.65}{3.69}\,\mathrm{m/s^2} = 3.97\,\mathrm{m/s^2}\) The acceleration of the blocks is \(3.97\,\mathrm{m/s^2}\) downwards on the inclined plane.