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March 4, 2015

March 4, 2015

Posted by **Matt** on Monday, March 12, 2007 at 4:53pm.

2nd question: A block of m = 2.00kg is moving a long a frictionless horizontal surface. The block reaches the relaxed spring (x ini. = 0) with an initial velocity of 6 m/s and compresses it to x final = 0.150m. At this value of compression, the speed of the block is reduced to v final = 4.00 m/s.

a) during this process, what is the change in Kinetic energy of the block?

sol) mv2(f)/2 - mv2 (i)/2 (correct?)

B) How much work has been done by the spring force on the block.

W(s)= mv2(i)/2 - mv2(f)/2

c) What is the spring constant of this spring. (hint: use work-kinetic energy theorem)

would it be x * whatever I get for part A?

Do any of the above answers make sense?

work done is negative when work is done on the system.

a correct

b correct

c No. change KE= 1/2 k x^2 solve for k.

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