Final answer:
The tension in the rope holding the 1.0 kg block is 3.038 N, which counteracts the frictional force due to kinetic friction. The acceleration of the 2.0 kg block is found to be 5.443 m/s² by deducting the frictional force from the applied tension and dividing by the mass of the block.
Step-by-step explanation:
To find the tension in the rope holding the 1.0 kg block to the wall, we first need to calculate the force of friction acting on the block due to kinetic friction. The force of kinetic friction (Ffriction) is given by Ffriction = μk * N, where N is the normal force and μk is the coefficient of kinetic friction.
For the 1.0 kg block, the normal force is equal to its weight, N = m * g, so the force of friction can be calculated as Ffriction = μk * m * g. With μk = 0.310, m = 1.0 kg, and g = 9.8 m/s2, the force of friction is Ffriction = 0.310 * 1.0 kg * 9.8 m/s2 = 3.038 N. This frictional force acts to the left, opposing the tension in the rope. Since the block is stationary, the tension must equal the force of friction, so the tension in the rope is 3.038 N.
To determine the acceleration of the 2.0 kg block, we use Newton's second law, ΣF = m * a, where ΣF is the sum of forces, m is the mass, and a is the acceleration. The force of kinetic friction on the 2.0 kg block is Ffriction,2 = μk * N2, where N2 is the normal force on the 2.0 kg block. Since the block has both the 1.0 kg block on top of it and its own weight to support, N2 = (m + m1) * g. Therefore, Ffriction,2 = 0.310 * (2.0 kg + 1.0 kg) * 9.8 m/s2 = 9.114 N.
The net force acting on the 2.0 kg block is the applied tension minus the frictional force, which is 20 N - 9.114 N = 10.886 N. We can now solve for acceleration, a = ΣF / m, giving us a = 10.886 N / 2.0 kg = 5.443 m/s2.