A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

We need to consider the forces acting on both the block and the cart. There will be gravitational forces acting on both Mass 1 and Mass 2, the force of static friction acting on M1, normal force acting between M1 and M2, and the externally applied force F acting on M2.

We will use Newton's second law (\(F = ma\)) to set up the equations for the forces acting on M1. The forces acting on M1 are: 1. Gravitational force: \(F_{g1} = M_1 * g\) 2. Normal force acting between M1 and M2: \(F_{N1}\) 3. Force of static friction: \(F_{f1} = \mu_s * F_{N1}\) The equation for the horizontal forces (x-direction) on M1 can be written as: \(F_{f1} = M_1 * a_1\), where \(a_1\) is the acceleration of Mass 1.

Similarly, using Newton's second law (\(F = ma\)), we can setup equations for the forces acting on M2. The forces acting on M2 are: 1. Gravitational force: \(F_{g2} = M_2 * g\) 2. Normal force acting between M1 and M2: \(F_{N2} = - F_{N1}\) 3. Force of static friction: \(F_{f2} = - F_{f1}\) 4. Externally applied force: \(F\) The equation for the horizontal forces (x-direction) on M2 can be written as: \(F - F_{f2} = M_2 * a_2\), where \(a_2\) is the acceleration of Mass 2.

To ensure that the block doesn't slip on the cart, the force of static friction should be equal to the product of the coefficient of static friction and the normal force, i.e., \(F_{f1} = \mu_s * F_{N1}\) Now we can substitute \(F_{f1}\) and \(F_{N1}\) in the equations of step 2 and step 3. Using these equations, we get: \(M_1 * a_1 = \mu_s * F_{N1}\) \(F - \mu_s * F_{N1} = M_2 * a_2\) Since M1 and M2 have the same acceleration, \(a_1 = a_2 = a\). We can solve these equations for F: \(F = (M_1 + \mu_s * M_2) * a\) Now we need to find the maximum value of F that prevents the slipping of the block. We can use the condition of maximum static friction: \(F_{f1} = \mu_s * F_{N1} = M_1 * a\) From the above equation, we can find the acceleration 'a': \(a = \dfrac{\mu_s * F_{N1}}{M_1}\) Substitute this value of acceleration back into the equation for F: \(F_{max} = (M_1 + \mu_s * M_2) * \dfrac{\mu_s * F_{N1}}{M_1}\) Now we need to find the normal force \(F_{N1}\). Since the system is in equilibrium vertically, we have: \(F_{N1} = M_1 * g\) Plug the values of \(M_1\), \(\mu_s\), \(M_2\), and \(g\) into the equation for \(F_{max}\): \(F_{max} = (0.640 + 0.620 * 0.320) * \dfrac{0.620 * (0.640 * 9.81)}{0.640}\)

Now, we can calculate the maximum force F that prevents the sliding of the block: \(F_{max} = 3.95308 \, N\) The maximum value of F that allows the block to accelerate with the cart, without sliding on top of the cart, is approximately 3.95 N.