Physics 101 Learning Object 1 Energy Conservation

1. Physics 101 Learning Object 1
Energy Conservation
A 5­kg mass spring system is held in the current position of the
figure below. Given that the total energy of the system is 20J and
the spring constant is 2.50 N/m (ignore friction),
a) What is the current kinetic energy of the system?
b) What is the Amplitude(A) of the system?
The mass is then released. At the point x1, the mass
experiences a velocity of 1.80m/s.
c) What is the current total energy of the system at x1?
d) What is the elastic potential energy of the system at x1?

2.
Solutions
a) When the mass is held at x = A, all energy in the mass
is stored as elastic potential energy. Thus, the kinetic
energy of the system at the time is zero.
b) Since all the energy is elastic potential energy, we can
use the equation E = (½)kA2
to solve for our amplitude:
E = (½) k A2
(20J) = (½) (2.50 N/m) A2
A = √(20J / (½) (2.50 N/m)
A = 4.00 m
c) After the mass is released, the total energy of the
system remains at 20J. The energy of a system is
always conserved, thus it remains constant.
d) When the system is moving, energy is continuously
transformed between kinetic and elastic potential
energy, but overall energy is conserved. So, we
can use E = K + U to solve our problem:
E = K + U
K = (½) m v2
E = ((½) m v2
) + U
(20J) = ((½)(5 kg)(1.8m/s)2
) + U
U = 20J ­ ((½)(5 kg)(1.80m/s)2
)
U = 20J ­ 8.10J
U = 11.9J