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Topic 3.4: Conservation of Mechanical Energy

Name: _______________________________ Date: _______________ Period: _______

Before We Start: Yesterday's Recap

A spring's potential energy is $\tfrac12kx^2$ and gravitational potential energy changes by $mg\Delta y$. A ball rolls off a table and falls. Name the two forms of energy the ball–Earth system has at the top, and the form it mostly has just before landing. What do you think stays the same the whole time?

Today's New Concepts

AP CED Alignment: Unit 3, Topic 3.4

Mechanical Energy and When It's Conserved

  • Total mechanical energy of a system: $E=K+U$ (with $U$ including every potential energy in that system)
  • Conserved ($E_i=E_f$, i.e., $K_i+U_i=K_f+U_f$) only when (1) no external force does work on the system AND (2) no energy is converted to internal energy
  • Energy changes form (bars change color); the total (bar height) stays the same

Worked Example: A $2.0\text{ kg}$ ball is released from rest $5.0\text{ m}$ above the bottom of a frictionless track. Total energy $E=mgh=100\text{ J}$. At a point $3.0\text{ m}$ above the bottom: $U=60\text{ J}$, $K=40\text{ J}$, $v=\sqrt{40}\approx6.3\text{ m/s}$. At the bottom: $K=100\text{ J}$, $v=10\text{ m/s}$.

Choosing the System

  • Forces between things in the system are internal (their energy is a $U$ on your list); forces from outside are external and can do work on the system
  • Block–Earth system: gravity is internal, so $U_g$ counts. Block alone: gravity is external, there is no $U_g$, and $\Delta K=W_{gravity}$
  • Block at rest on a vertical spring — block–spring–Earth: $U_g$ + $U_s$. Block–spring: only $U_s$

Worked Example: A spring–block system (spring $k=200\text{ N/m}$ compressed $0.10\text{ m}$, block $0.50\text{ kg}$, frictionless surface) has $E=U_s=1.0\text{ J}$ at the start. At the relaxed length it's all kinetic: $v=\sqrt{2(1.0)/0.50}=2.0\text{ m/s}$.

Solving With Conservation

  • The mass cancels in gravitational problems, and the path's shape never matters — only the start and end heights and speeds
  • Going up a height $h$: $v_f=\sqrt{v_i^2-2gh}$. Going down: $v_f=\sqrt{v_i^2+2gh}$

Worked Example: A sled at $8.0\text{ m/s}$ climbs a frictionless hill $2.0\text{ m}$ high: $v_f=\sqrt{64-2(10)(2.0)}=\sqrt{24}\approx4.9\text{ m/s}$.