Question

As shown in the figure with Problem 4.94 , a block of mass \(M_{1}=0.250 \mathrm{~kg}\) is initially at rest on a slab of mass \(M_{2}=0.420 \mathrm{~kg}\), and the slab is initially at rest on a level table. A string of negligible mass is connected to the slab, runs over a frictionless pulley on the edge of the table, and is attached to a hanging mass \(M_{3}=1.80 \mathrm{~kg} .\) The block rests on the slab but is not tied to the string, so friction provides the only horizontal force on the block. The slab has a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.340\) with both the table and the block. When released, \(M_{3}\) pulls on the string, which accelerates the slab so quickly that the block starts to slide on the slab. Before the block slides off the top of the slab: a) Find the magnitude of the acceleration of the block. b) Find the magnitude of the acceleration of the slab.

Short answer

Answer: The magnitude of the acceleration of the block, M1, is approximately 1.12 m/s². The magnitude of the acceleration of the slab, M2, is approximately 3.86 m/s².

Step-by-step solution

Analyze forces acting on the system

Before we begin, we need to understand the forces involved in this problem. On \(M_{1}\), we have the gravitational force (\(M_{1}g\)) and the friction force (\(f_{1}\)) that acts between the block and the slab. On \(M_{2}\), we have the gravitational force (\(M_{2}g\)), the tension in the rope (\(T\)), and the friction force between the slab and the table (\(f_{2}\)). On \(M_{3}\), we have the gravitational force (\(M_{3}g\)) and tension in the rope (\(T\)).

Write expressions for the frictional forces

We are given the coefficient of kinetic friction \(\mu_{k}\) between the block and the slab, and between the slab and the table. We can use this to write expressions for the frictional forces involved: \(f_1 = \mu_{k} M_1g\) \(f_2 = \mu_{k} M_2g\)

Apply Newton's second law to M1 and M2

Using Newton's second law (\(F=ma\)) for both \(M_1\) and \(M_2\), we can find the acceleration of each mass. Let's call the acceleration of \(M_1\) as \(a_1\) and the acceleration of \(M_2\) as \(a_2\). For \(M_1\): \(a_1 = \frac{f_1}{M_1}\) For \(M_2\): \(a_2 = \frac{T - f_2}{M_2}\)

Apply Newton's second law to M3

Now we can apply Newton's second law to \(M_3\). In this case, the only forces acting on \(M_3\) are the tension in the rope and the gravitational force. We can write an equation for the acceleration in terms of tension: \(T = M_3g - M_3a_2\)

Solve for the acceleration of the block (M1) and the slab (M2)

First, let's find the accelerations of the block, \(M_1\). We can use the expression for the friction force \(f_1\): \(a_1 = \frac{\mu_{k} M_1g}{M_1}\) Substitute the given values of \(\mu_{k} = 0.340\) and \(M_1 = 0.250 \, kg\), and solve for \(a_1\): \(a_1 = \frac{0.340 \times 0.250 \times 9.81}{0.250} \approx 1.12 \, m/s^2\) Now, we can solve for the acceleration of the slab, \(M_2\). First, we substitute the expression for \(T\) that we found in step 4 into the equation for \(a_2\): \(a_2 = \frac{M_3g - M_3a_2 - f_2}{M_2}\) Now, substitute the given values of \(M_3 = 1.80 \, kg\), \(M_2 = 0.420 \, kg\), and the expression for \(f_2\): \(a_2 = \frac{1.80 \times 9.81 - 1.80a_2 - 0.340 \times 0.420 \times 9.81}{0.420}\) This is an equation with one unknown variable, \(a_2\). We can solve for \(a_2\): \(a_2 \approx 3.86 \, m/s^2\)

Results

a) The magnitude of the acceleration of the block, \(M_1\), is approximately \(1.12 \, m/s^2\). b) The magnitude of the acceleration of the slab, \(M_2\), is approximately \(3.86 \, m/s^2\).