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.

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yudhvir maan · Answer

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

Ashwani Goel · Answer

ans.. is 1.8mm

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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/s 2 ) and the mass less spring constant is 50 N/m. Find the distance stretched of the spring.

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.

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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/s 2 ) and the mass less spring constant is 50 N/m. Find the distance stretched of the spring.

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.

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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.