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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:
$E_{f} = E_{i} + W$
$K_{f} + U_{f} = K_{i} + U_{i} + W$
$K_{f}$ and $W$ cancel out.
$0 + \dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}mv^2_{i} + \dfrac{1}{2}k_{s}s^2_{i}$
$\dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}(0.2 kg)(0.5 m/s)^2 + \dfrac{1}{2}(12 N/m)(0.03 m)^2$
$\dfrac{1}{2}k_{s}s^2_{f} = 0.03 J$
$s_{f} = \sqrt {2(0.03 J)/(12 N/m)} = 0.07 m$
$s_{f} = 0.07 m$
b:
$K_f + 0 = 0.03 J$
$\dfrac{1}{2}mv^2_{f} = 0.03 J$
$v_{max} = \sqrt {2(0.03 J)/(0.2 kg)} = 0.55 m/s$