Question

Two blocks \(\left(m_{1}=1.23 \mathrm{~kg}\right.\) and \(m_{2}=2.46 \mathrm{~kg}\) ) are glued together and are moving downward on an inclined plane having an angle of \(40.0^{\circ}\) with respect to the horizontal. Both blocks are lying flat on the surface of the inclined plane. The coefficients of kinetic friction are 0.23 for \(m_{1}\) and 0.35 for \(m_{2}\). What is the acceleration of the blocks?

Short answer

Answer: The acceleration of the blocks is approximately 1.845 m/s².

Step-by-step solution

Identify Forces Acting on the Blocks

We have three main forces acting on the blocks: 1. Gravitational Force: This force is acting downwards and is equal to \(m_{1} g\) for block 1 and \(m_{2} g\) for block 2, where \(g\) is the acceleration due to gravity (9.81 \(m/s^2\)). 2. Normal Force: This force is acting perpendicular to the inclined plane and is equal to \(m_{1} g \cos(40^{\circ})\) for block 1 and \(m_{2} g \cos(40^{\circ})\) for block 2. 3. Frictional Force: This force is acting up the inclined plane (opposite to the direction of motion) and is equal to \(\mu_{1} N_{1}\) for block 1 and \(\mu_{2} N_{2}\) for block 2, where \(\mu_{1}\) and \(\mu_{2}\) are the coefficients of kinetic friction, and \(N_{1}\) and \(N_{2}\) are the normal forces on block 1 and block 2 respectively.

Calculate the Net Force along the Incline

Now we will calculate the net force acting on the blocks along the incline. The gravitational force has two components: one parallel to the inclined plane (\(F_{g_{\parallel}}\)) and one perpendicular to the inclined plane (\(F_{g_{\perp}}\)). For block 1: \(F_{g_{1 \parallel}} = m_{1} g \sin(40^{\circ})\) For block 2: \(F_{g_{2 \parallel}} = m_{2} g \sin(40^{\circ})\) The net force acting along the incline can be calculated as: \(F_{net} = F_{g_{1 \parallel}} + F_{g_{2 \parallel}} - F_{1f} - F_{2f}\) Substituting the known values, we get: \(F_{net} = 1.23g \sin(40^{\circ}) + 2.46g \sin(40^{\circ}) - 0.23(1.23g \cos(40^{\circ})) - 0.35(2.46g \cos(40^{\circ}))\)

Use Newton's Second Law to Calculate Acceleration

Now we will use Newton's second law (\(F = ma\)) to calculate the acceleration of the blocks. We will consider the blocks as a single system of mass \(M = m_{1} + m_{2}\). \(F_{net} = Ma\) Rearranging and plugging in the expression for net force, we get: \(a = \frac{1.23g \sin(40^{\circ}) + 2.46g \sin(40^{\circ}) - 0.23(1.23g \cos(40^{\circ})) - 0.35(2.46g \cos(40^{\circ}))}{1.23 + 2.46}\) Finally, substituting the value of g (9.81 \(m/s^2\)) and evaluating the expression, we find the acceleration of the blocks: \(a \approx 1.845 \:\mathrm{m/s^2}\)