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The force for a spring (facing up an incline) pushes with a force proportional to the distance from its rest position.Lets say that the 2kg block is pushed up a 30 degree incline.The change in the spring's potential energy (from the maximum compression to the release point/ aka natural unstretched position) is 39.2 JoulesMy book is saying to plug in 39.2 Joules into the 1/2 m v squared equation.The book says this is how to solve for the velocity at the spring's unstretched natural position.My question is (AM I WRONG OR IS MY BOOK WRONG) what about the work done by gravity during the time while my spring was doing work?The 2 kg object actually moves up a 30 degree incline (thus the sine of 30 is 0.5) so every meter of displacement WHILE the spring is imputing work into the kinetic energy of the object gravity is also imputing negative work (opposite the direction of motion) given by the formula mgh.My book solution's guide doesn't even mention the mgh work?The ONLY creative explanation I have is that maybe the spring continues to impute work into the object slightly beyond the x=0 position (unstretched relaxed state of spring) because at x=0 the force from the spring is zero but gravity is pushing "into the spring" and the spring will thus be in contact "slightly long" than the rest-position of the spring.Is this the solution? Did I outthink my physics book? Does the spring do work beyond x=0? Does that work (done beyond the spring's rest position) precisely equal and net-out to zero the amount of negative work done by the mgh equation which acted on it during the event?Help, thanks
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