The 0.18 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from th
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Answer:
Please refer to the attached schematic for free-body diagrams of the blocks.
The contact forces N1 = mg, N2 = 3mg.
The friction force fs = mg/2.
The tension force T = 3mg/2.
The distance y = (gt^2)/4
Explanation:
Starting with the third block which is declining a time t. Newton's Second Law implies that
There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.
Let's look at free-body diagram 1:
Newton's Second Law:
That is another equation with one more unknown: fs.
Free-body diagram 2:
That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:
Plugging this equation into the first equation gives
Following this, we can find the other unknowns:
The contact forces are N1 and N2.
Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.
Answer:
The final answer is
Explanation:
Hence,