posted by leeece on .
A block of mass = 0.85 kg is connected to a spring of force constant = 775N/m on a smooth, horizontal surface.
1) Plot the potential energy of the spring from = -5.00 cm to = 5.00 cm.
2) Determine the turning points of the block if its speed at x = 0 is 1.1 m/s
I am so stumped on this problem. Could someone please help?
To plot the potential energy, simply plug your givens into the equation U=(1/2)kx^2. You should get a parabola that goes to 0 in the center.
To determine the turning points of the block, use the law of conservation of energy. You know that Ui+Ki = Uf+Kf since there is no friction. At x=0, there will be no potential energy (Ui=0). A "turning point" is where the motion will stop and the block will move in the opposite direction. This means that Kf=0.
So, set Ki=Uf
(1/2)mv^2 = (1/2)kx^2
(1/2)(.9)(1.2)^2 = (1/2)(775)x^2
Solve for x=.0409 m
This will be the first turning point. The other turning point will be the opposite of this (remember the parabola) - so x=-.0409 m.