Posted by Jimmy on Saturday, June 2, 2012 at 8:13pm.
a) ME= 1/2 k x^2
b) 1/2 m v^2=change in PE= 1/2 k (.27^2-.09^2)
In turns out that for b), instead of using
(.27^2 -0.09^2), we have to use (.27^2-0.18^2).
I think your solution tells us when the spring moves 18 cm, not 9 cm.
Thank you, though. This was really useful.
Related Questions
Physics - A 0.495-kg mass is attached to a horizontal spring with k = 108 N/m. ...
Physics - A 0.51-kg mass is attached to a horizontal spring with k = 108 N/m. ...
physics - A new type of spring is found where the spring force F is given by F=-...
Physics - A 0.30 kg mass is attached to a spring with a spring constant 170 N/m ...
Physics - Block A of mass 2.0 kg and Block B of 8.0 kg are connected by a spring...
Physics - A 2.50 mass is pushed against a horizontal spring of force constant 26...
physics - a 0.75 kg mass attached to avertical spring streches the spring 0.30m...
Physics - A 2.5 m rod of mass M= 4. kg is attached to a pivot .8m from the left ...
Physics - Need help on part B! A 2.50 mass is pushed against a horizontal spring...
physics - a 0.1 kg mass is suspended at rest from a spring near the Earth's...
For Further Reading