Question #152337

A 2.00 kg block is pushed against a spring with negligible mass and stiffness k=400N/m, compressing it by 0.220 m.
When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37o (see figure below). 



1
Expert's answer
2020-12-21T11:00:14-0500

Answer

Maximum height attain by block on inclined plain can be given as

12Kx2=mgh\frac{1}{2}Kx^2=mgh


By putting the value

12(400)(0.220)2=2×9.8×hh=0.5m\frac{1}{2}(400) (0.220) ^2=2\times 9.8\times h\\h=0.5m


So maximum height attain by block on inclined plane is 0.5m.




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