I am looking for the answer key to problem 43 , chapter 9, from Text book Matter
ID: 2253525 • Letter: I
Question
I am looking for the answer key to problem 43 , chapter 9, from Text book Matters and Interaction, 3rd edition, by Ruth Chabay
Moving blocks connected by a spring
Two identical 0.15 kg blocks (labeled 1 and 2) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper diagram, where x2 = 0.04 m. Then a constant force of 6 N to the right is applied to block 2, and at a later time the blocks are in the new positions shown in the lower diagram, where x1 = 0.01 m and x3 = 0.1 m. At this final time, the system is moving to the right and also vibrating, and the spring is stretched.
Explanation / Answer
When the spring is compressed, the CM is at [ 0.04 - 0.01 ] / 2 = 0.025
When the spring is completely stretched and when completely, difference in position of second block = 0.099 - 0.04 = 0.059
Thus, the second block would be at 0.0475 from the first block at equilibium.
When the spring is stretched, the PE stored = 1/2 * k.x^2 = 1/2 * (6/x) * x^2 .
Here, x = 0.0295.
So, PE = 0.0885 J.
In this case and all cases involving 2 blocks connected by spring, the CM moves at a constant speed, and the body will vibrate about this centre of mass (CM).
SORRY, I COULDNT EXPLAIN MORE.
THIS IS THE CONCEPT. HOPE YOU SOLVE IT.
THANK YOU. DO RATE.
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