The spring is already stretched by the weight before is is pulled. That amount of "equilibrium" stretch is
Xe = M g/k = 0.674 m. Then the spring is pulled downward another 0.200 m for a total stretch of X = 0.874 m at the point of release. The kinetic energy is zero at that point, and the spring potential energy is (1/2)kX^2 = 18.33 J . Relative to the the point of release, the position is then h = 0. Call the gravitational potential energy zero at that point also.
The total energy of the system is the spring energy of 18.33 J
At h = 0.2, X = 0.674 m
At h = 0.4, X = 0.474 m
gravitational PE (gPE) = M g h
spring potential energy = (1/2) k X^2
kinetic energy = 18.33 J - KE - gPE
You complete the table
A 3.30 kg object is hanging from the end of a vertical spring. The spring constant is 48.0 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.
Find KE, PE-gravity, PE-elastic and E for each h (0, 0.200, 0.400)
1 answer
1 answer