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Grade 12th passIIT JEE Entrance Exam

a 263g block is dropped on to a vertical spring with force constant k=2.52N/cm as shown in the below figure the block sticks to the spring ,and the spring compresses 11.8cm before coming momentarily to rest .while the spring is being compressed ,how much work is done (a)by the force of gravity and (b) by the spring © what was the speed of the block just before it hit the spring ?(d)If the initial speed of the block is doubled,what is the maximum compression of the spring?Ignore friction.

Profile image of sai
8 Years agoGrade 12th pass
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1 Answer

Profile image of Arun
8 Years ago
Ke of block = (1/2) m v2
when spring is uncompressed
(a) work done by gravity on mass is m g h
= .263*9.81*.118 Joules
(b) work done by spring on mass is -(1/2)kx2 = -.5(252 N/m)(.118)2
(c) so because v = 0 at bottom
initial Ke + work done by g = work done on spring by mass
.5 (.263)v2 + .263*9.81*.118 = .5(252 N/m)(.118)2
solve for v