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(a) Calculate the speed of the glider at the point where it has moved 0.180 m from its starting point, so that the spring is momentarily exerting no force.

So potential = kinetic energy

1/2kx^2=1/2mv^2

v^2=kx^2/m

v^2=(9)(.18^2)/.15

v=1.39

(b) Calculate the speed of the glider at the point where it has moved 0.250 m from its starting point.

Here x=.07 because isnt x the distance from the equilibrium point, which is at .18? If so, use the same equation

1/2kx^2=1/2mv^2

v^2=(9)(.07^2)/.15

v=.542 m/s

Thanks for checking