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For the following Fig. M=2 Kg and m=1kg . The horizontal surface is frictionless and the group move with constant acceleration (3.27 m/s2) and the mass less spring constant is 50 N/m. Find the distance stretched of the spring.

1033_24554_Figure

Reyad Khaled , 15 Years ago
Grade Upto college level
anser 2 Answers
yudhvir maan

FOR MASS m

mG-T=ma therefore T=6.73 N

FOR MASS M

F=Ma here mass =2kg and a=3.27

therefore F=6.54 N

F=kx where k=50 and x is elongation

x=6.54/50meter

Last Activity: 15 Years ago
Ashwani Goel

ans.. is     1.8mm 


Last Activity: 15 Years ago
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