Final answer:
The total distance covered by the block on the horizontal surface before coming to rest is 64 m.
Step-by-step explanation:
To determine the total distance covered by the block on the horizontal surface before coming to rest, we need to analyze the motion on each path separately.
First, on path AB, the surface is smooth, meaning there is no friction. Therefore, the block will continue its motion with a constant velocity. We can use the equation v² = u² + 2as to calculate the distance covered on path AB, where v is the final velocity (which is 0 m/s in this case), u is the initial velocity (which is the speed of the block at point B), a is the acceleration (which is 0 m/s² as there is no friction), and s is the distance covered. Rearranging the equation, we have s = (v² - u²) / (2a). Substituting the values, we get s = (0 - (8 m/s)²) / (2 * 0 m/s²) = -32 m. Since distance cannot be negative, we take the magnitude, so the distance covered on path AB is 32 m.
Next, on path BC, there is a coefficient of friction of μ = 0.1. The frictional force can be calculated using the equation f = μN, where f is the frictional force, μ is the coefficient of friction, and N is the normal force. The normal force is equal to the weight of the block, which is N = mg, where m is the mass of the block and g is the acceleration due to gravity. Substituting the values, we have N = (0.1 kg)(10 m/s²) = 1 N. Therefore, the frictional force is f = (0.1)(1 N) = 0.1 N. Since the block is in perfectely elastic collision with the vertical wall at C, it will rebound with the same velocity but in the opposite direction. This means that the block will cover the same distance on path BC as it did on path AB, which is 32 m.
Therefore, the total distance covered by the block on the horizontal surface before coming to rest is the sum of the distances covered on paths AB and BC, which is 32 m + 32 m = 64 m.