Presentation problem #7: Work and Energy Ground rules One designated person (the

Question

Presentation problem #7: Work and Energy Ground rules One designated person (the "presenter") within the group will be responsible for wrising up the group's help) a self-contained, succinct outline of the solution method and the problem's solution 2. Large (I 17, ledger) she eets will be provided, make sure your lettering/diagrams are large enough and clear enough that others can follow your solution at more than an arm's length away and Be sure your answer includes appropriate units, and that vector quantities direction (or components). A sketch or diagram will almost always be required, sometimes a graph may make things clearer 3. Presentation write-ups are due at the end of the lab session during which they were assigned Each solution write-up will be scored (from zero to ten, using the laboratory grading scale). member of the group receives thi presenter), and the clear and careful as possible), Presenters will be rotated within the assigned lab groups so that each student has the 4. t s score (so it is in each contributor' s interest to assist the presenter receives twice this score (so it is in the presenter's interest to be as opportunity to serve as presenter at least three times. 5. These scores will also be added, and divided by the total number of available presentation points (typically 120); that fraction will determine 20 points of the overall lab grade. Group B A 2.0 kg block is dropped from a height of 48 cm onto a massless plate supported by a spring (k - 1600 /m). At the point of maximum compression of the spring, the block cracks into two equal pieces, one of which falls off the plate. How high in the air above the plate is the second piece subsequently propelled?

Explanation / Answer

Applying energy conservation to find maximum compression,

m g (h + d) - k d^2 /2 = 0 - 0

k d^2 /2 - m g d - m g h = 0

800 d^2 - 19.6 d - 9.42= 0

d = 0.121 m

now m = 1 kg

Applying energy conservation again,

1 x 9.81 x (0.121 + h) - 800(0.121^2)= 0

h = 1.07 m

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