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Grade 12Mechanics

The ends of a spring are attached to masses 2kg and 3kg. the 3kg block rests on a horizontal surface and 2kg block which is vertically above it is in eqiulibrium producing compression of 1cm of spring initially. find minimum value of distance the 2kg mass must be displaced down so that when it is released the 3kg block may lift off from the ground.
please answer fast.

Profile image of Kuldip Jha
10 Years agoGrade 12
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1 Answer

Profile image of Swaraj Prateek
10 Years ago
At equillibrium
20=Kx1, given x1=0.01m
K=2000 N/cm
Now to lift of 3Kg mass 50N force required.
So Kx2=50N
x2=50/2000=0.025m
Further compression required x2-x1=2.5-1=1.5cm
Answer is 1.5cm