A 42 -kg slab rests on a frictionless floor. A \(9.7-\mathrm{kg}\) block rests on top of the slab, as in Fig. 5-40. The coefficient of static friction between the block and the slab is \(0.53\), while the coefficient of kinetic friction is \(0.38\). The \(9.7-\mathrm{kg}\) block is acted on by a horizontal force of \(110 \mathrm{~N}\). What are the resulting accelerations of \((a)\) the block and \((b)\) the slab?

The maximum static frictional force the slab can exert on the block is given by \(f_{s_{max}} = \mu_s \cdot N\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force. The normal force is equal to the weight of the block, which is its mass times gravity. Hence, \(N = m \cdot g\), with \(m = 9.7 kg\) and \(g = 9.8 m/s^2\). Hence, \(f_{s_{max}} = 0.53 \cdot 9.7 kg \cdot 9.8 m/s^2 = 51.0266 N\). If the applied force is greater than this maximum static frictional force, the block will slip relative to the slab. Since the applied force of \(110 N\) is greater than \(f_{s_{max}}\), the block slips.

Now we consider the forces acting on the block and the slab separately. The forces acting on the block are its weight, the normal force from the slab, the applied force, and the force of kinetic friction from the slab since the block is slipping. The kinetic frictional force is given by \(f_k = \mu_k \cdot N\), where \(\mu_k\) is the coefficient of kinetic friction, yielding \(f_k = 0.38 \cdot 9.7 kg \cdot 9.8 m/s^2 = 36.1456 N\). The net force acting on the block is the applied force minus the frictional force, or \(110 N - 36.1456 N = 73.8544 N\). According to Newton's second law, this net force will cause the block to accelerate, with the acceleration given by \(a = F/m\), or \(a_block = 73.8544 N / 9.7 kg = 7.61 m/s^2\).

For the slab, the only horizontal force acting upon it is the kinetic frictional force exerted by the block, which will cause the slab to accelerate. Hence, the net force acting on the slab is the frictional force, or \(36.1456 N\). The acceleration of the slab is then \(a_slab = F/m = 36.1456 N / 42 kg = 0.86 m/s^2\).