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Example: Thermal Equilibrium - Part A
A metal block of mass 3 kg is moving downward with speed 2 m/s when the bottom of the block is 0.8 m above the floor. When the bottom of the block is 0.4m above the floor, it strikes the top of a relaxed vertical spring 0.4 m above the floor, it strikes the top of a relaxed vertical spring 0.4 m in length. The stiffness of the spring is 2000 N/m.
a: The block continues downward, compressing the spring. When the bottom of the block is 0.3m above the floor, what is its speed?
b: The block eventually heads back upward, loses contact with the spring, and continues upward. What is the maximum height reached by the bottom of the block above the floor?
c: What approximations did you make?
Facts
a:
Initial State: Block 0.8 m above floor, moving downward, vi = 2 m/s, spring relaxed
Final State: Block 0.3 m above floor, spring compressed
b:
Initial State: Block 0.8 m above floor, moving downward, vi = 2 m/s, spring relaxed
Final State: Block at highest point, vf=0, spring relaxed
Lacking
a: The speed of the block when it is 0.3 m above the floor
b: The maximum height
Approximations & Assumptions
Air resistance and dissipation in the spring are negligible.
Ug≈mgy near Earth's surface.
ΔKEarth is negligible.
Representations
System: Earth, block, spring
Surroundings: Nothing significant
Solution
a:
From the Energy Principle:
Ef=Ei+W
Kf+Us,f+Ug,f=Ki+Us,i+Ug,i+W
Us,i and W cancel out.
12mv2f+mgyf+12kss2f=mgyi+12mv2i
v2f=v2i+2g(yi−yf)−ksms2f
vf=√(2m/s)2+2(9.8N/kg)(0.5m)−2000N/m3kg0.1m2
b: