A 1.90-kg object is hanging from the end of a vertical spring. The spring consta
ID: 1977891 • Letter: A
Question
A 1.90-kg object is hanging from the end of a vertical spring. The spring constant is 37.0 N/m. The object is pulled 0.200 m downward and released from rest. Complete the following table 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 listed. The vertical positions h indicate distances above the point of release, where h = 0 m.
h (m) KE (J) PEgravity (J) PEelastic (J) E (J) 0 0.200Explanation / Answer
initial position kx=mg 37*x=1.9*9.8 x=0.5 object is pulled 0.2 upward then x=0.7 PE gravity=0 (assuming 0.7 is refrence point) PE elastic=kx^2/2=37*(0.7)^2/2=9.065 KE=0 velocity is 0. at x=0.5 or h=0.2 PE gravity =1.9*9.8*0.2=3.724=mgh PE elastic=kx^2/2=37*(0.5)^2/2=4.625 KE =9.065-(3.724+4.625)=0.716 (using energy conservation between h=0 and h=0.2 total energy(PE gravity+PE elastic+KE) should be same between two points)
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