A 1.50 kg object is hanging from the end of a vertical spring. The spring consta
ID: 2070033 • Letter: A
Question
A 1.50 kg object is hanging from the end of a vertical spring. The spring constant is 60.0 N/m. The object is pulled 0.200 m downward and released from rest. Complete the table below by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions indicated. The vertical positions h indicate distances above the point of release, where h = 0.h (m) KE (J) PE-gravity (J) PE-elastic (J) E (J)
0 0 0
0.200 1.2
0.400 0
Explanation / Answer
kx= mg 60*x= 15 x= .25m x'=.45m= h=0 at h=0 KE=0, PE=0, PE(elastic)=1/2 kx'2= energy of the system 1/2*60*.45^2= 6.075 J at h=.2m PE= mgh= 15*.2= 3J PE(elastic)=1/2 kx'2= 1/2*60*(.45-.2)^2=1.875J KE= 6.075-3-1.875=1.2 total energy= 6.075J at h=.4m PE= mgh= 15*.4=6J PE(elastic)=1/2 kx'2= 1/2*60*.05^2=0.075 KE=0 total energy= 6.075J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.