A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide freely along the smooth vertical guides BC and
DE. Determine the velocity of the block and plate just after impact. The coefficient of restitution between the block and the plate is e = 0.75. Also, find the maximum compression of the spring due to impact.The spring has an unstretched length of 600 mm
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate
1 answer:
Answer:
Check the explanation
Explanation:
<u>Given data </u>
Mass of the block A is Ma = 10kg
Mass of the plate is Mp = 5kg
Coefficient of restitution between the block and that of the plate is e=0.75
Unstretched length of the spring is 0.6m
From the conservation of energy, we have four motion of block A from (1) to (2)
Then accordingly, we will have the step by step calculation in the attached image below .
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4.75cm * 5.22cm is 24.795cm²
but let's do this with m² instead:
0.0475m * 0.0522m = 0.0024795
now we can compare it with the 49780 much easier.
devide. that's the <u>roughly</u> 20,000,000fold of the plot, but it's still squared, so let's take the root
so the representative fraction is 1:4,480.
the exact value is in the screens
have a nice day:)
