Well, it sounds like we have a classic physics problem on our hands! But don't worry, I'll do my best to clown around and simplify it for you.
First, we need to find the acceleration of the system. Since the two blocks are connected by a rope passing over a pulley, they will have the same acceleration. Let's call this acceleration "a".
To find "a", we can use the following equation: ΣF = (M1 + M2) * a, where ΣF is the sum of the forces acting on the system.
The forces acting on the system are the gravitational forces pulling the blocks down. For M1, we have a force of M1 * g (mass of M1 times the acceleration due to gravity). And for M2, the force is M2 * g.
Now, let's plug in the given values. We have M1 = 2 kg, M2 = 4 kg, and g ≈ 9.8 m/s².
For M1: F1 = M1 * g = 2 kg * 9.8 m/s² = 19.6 N
For M2: F2 = M2 * g = 4 kg * 9.8 m/s² = 39.2 N
So, the sum of the forces is ΣF = F2 - F1 = 39.2 N - 19.6 N = 19.6 N.
Now, let's plug this value into the equation:
19.6 N = (2 kg + 4 kg) * a
Simplifying:
19.6 N = 6 kg * a
a ≈ 3.27 m/s²
Now that we have the acceleration, let's figure out the time it takes for the 4 kg block to reach the floor from a height of 1 m.
We can use the following equation: Δd = V0 * t + 0.5 * a * t².
Since the 4 kg block is released from rest, its initial velocity (V0) is 0. And Δd is the distance it travels, which is 1 m.
1 m = 0 * t + 0.5 * 3.27 m/s² * t²
Rearranging the equation:
0.5 * 3.27 m/s² * t² = 1 m
Solving for t:
t² ≈ (1 m) / (0.5 * 3.27 m/s²) ≈ 0.306 s²
t ≈ √(0.306 s²) ≈ 0.553 s
So, it takes approximately 0.553 seconds for the 4 kg block to reach the floor from a height of 1 m.
I hope my clowning around made this problem a little less daunting for you!