A box of mass m=1.5 kg is attached to a spring with force constant k=12 and suspended on a frictionless incline that makes a 30 degree angle with respect to the horizontal. With the spring in its unstretched length, the box is released from rest at x=0. The box slides down the ramp as the spring stretches and momentarily comes to rest at x=xmax.
What is xmax, the maximum extension of the spring when the box is at its lowest point on the incline?
I know the answer is 1.23 but I can't figure out how to get it! I first tried using the forces, setting the force spring equal to the force due to gravity acting on the box, this gave me half of the answer. Should I be using potential energy? I keep getting the wrong answer no matter which equations I use, please help!
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What I did on this was mgsintheta = .5kx^2... kept getting it wrong. then realized you have to factor in the distance traveled by the block, that the spring is stretching out to, so what you get is mgsintheta""X"" = .5kx^2 or mgsintheta=.5kx
1 answer