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Example: Energy in a Spring-Mass System
A mass of 0.2 kg is attached to a horizontal spring whose stiffness is 12 N/m. Friction is negligible. At t = 0 the spring has a stretch of 3 cm and the mass has a speed of 0.5 m/s.
a: What is the amplitude (maximum stretch) of the oscillation?
b: What is the maximum speed of the block?
Facts
a:
Initial State: s_i = 3 cm, v_i = 0.5 m/s
Final State: v_f = 0 (maximum stretch)
b:
Initial State: s_i = 3 cm, v_i = 0.5 m/s
Final State: Maximum speed (s_f = 0)
Lacking
Approximations & Assumptions
Representations
System: Mass and spring
Surroundings: Earth, table, wall (neglect friction, air)
Solution
a:
From the Energy Principle:
Ef=Ei+W
Kf+Uf=Ki+Ui+W
Kf and W cancel out.
0+12kss2f=12mv2i+12kss2i
12kss2f=12(0.2kg)(0.5m/s)2+12(12N/m)(0.03m)2
12kss2f=0.03J
sf=√2(0.03J)/(12N/m)=0.07m
sf=0.07m
b:
Kf+0=0.03J
12mv2f=0.03J
vmax=√2(0.03J)/(0.2kg)=0.55m/s