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Lesson 8 of 16

Work, Energy and Power · Lesson 8 of 16

Friction, Dissipation and the Energy Balance

“Extend mechanical-energy accounting to friction, drag and external work.”

Learning Objectives

• Derive Δ(K+U) from work by remaining forces. • Separate mechanical-energy loss from total-energy loss. • Solve spring compression and drag problems with dissipation.

Where mechanical energy goes

A sliding block slows while its surface and the floor warm up. Its organised translational energy has become internal energy of the materials and surroundings. Energy has not vanished. When we track only kinetic and conservative potential energy, we deliberately leave out that internal-energy account. A complete system energy balance includes energy transfers through its boundary and changes of internal, chemical and other stored forms.

Mechanical energy with other workLaTeX
U includes the chosen conservative interactions. Wother includes all remaining work, such as an applied push, a motor or sliding friction.

Many textbooks write Wnc for this remaining work. A positive value adds mechanical energy; a negative value removes it. Do not count a conservative interaction twice by including both its potential-energy change and its work on the same side without the appropriate rearrangement. The work–energy theorem still holds even when mechanical energy is not conserved.

Sliding friction and stopping

A rough horizontal stop

Problem
A 2 kg block starts at 6 m/s on a level surface with μk = 0.3. Use g = 10 m/s². Find stopping distance.

  1. 1.Initial kinetic energy is ½×2×36 = 36 J. Friction magnitude is μmg = 6 N.
  2. 2.There is no change in gravitational potential energy. Friction work is −6d.
  3. 3.Set 0 − 36 = −6d to get d = 6 m. The materials gain 36 J of internal energy in the ideal isolated block-plus-surface account.

Maximum compression with friction

Rough-surface spring compressionLaTeX
This starts when the body touches an initially undeformed spring. Add friction work over any earlier rough approach separately.
Car and spring with resistance

Problem
Use m = 1000 kg, v0 = 5 m/s, k = 5250 N/m, μk = 0.5 and g = 10 m/s². Find maximum compression.

  1. 1.Normal force is mg, so sliding friction is 5000 N. During compression, both friction and spring force oppose motion.
  2. 2.The energy equation is 12500 = 2625x² + 5000x.
  3. 3.Solve 5250x² + 10000x − 25000 = 0. The positive root is x = 10/7 m ≈ 1.43 m.
  4. 4.At this x, spring energy is about 5357.14 J and friction has dissipated 7142.86 J; the sum is 12500 J.
  5. 5.This is less than the frictionless 2.18 m. The source’s printed 1.35 m does not satisfy the stated data; substituting it gives only 11533.13 J.

Drag and terminal motion

A falling water drop

Problem
A spherical water drop of radius 2 mm falls 500 m from rest and reaches the ground at 10 m/s. Use water density 1000 kg/m³ and g = 9.8 m/s². Find gravity work in each half and total drag work.

  1. 1.Mass is ρ(4πr³/3) = 3.351×10⁻⁵ kg.
  2. 2.Each 250 m half gives Wg = mg×250 ≈ 0.08210 J, regardless of how acceleration changes. Total gravity work is 0.16420 J.
  3. 3.Final K = ½m×10² ≈ 0.001676 J, hence Wdrag = ΔK − Wg,total ≈ −0.16252 J, approximately −0.163 J.
  4. 4.If terminal speed is reached halfway, K is constant during the second half. There, drag work exactly cancels gravity work; neither individual work is zero.

Damped pendulums and decaying orbits

A pendulum loses 5% of its initial energy

Problem
A 1.5 m pendulum is released from horizontal. Air resistance removes 5% of its initial mechanical energy before the bottom. Use g = 9.8 m/s². Find bottom speed.

  1. 1.The release height above the bottom is L. Initial energy is mgL with zero potential at the bottom.
  2. 2.Only 0.95mgL remains as kinetic energy there. Thus v = √(1.90gL) = √27.93 ≈ 5.28 m/s.
  3. 3.Subtract the loss from energy, not directly from speed.

A satellite can lose total mechanical energy to drag while speeding up as its orbit descends. Its gravitational potential becomes more negative; its kinetic energy can rise while their sum falls. In the slowly changing nearly circular approximation, K = GMm/(2r), U = −GMm/r and E = −GMm/(2r). Decreasing r increases K but decreases E. This comparison is between successive nearly circular orbits; drag itself locally acts against the velocity. A rocket casing heated by atmospheric resistance similarly draws energy from relative mechanical motion, with the distribution depending on the chosen system and frame.

Check your understanding

Quiz

Quick check

Friction reducing K means total energy

Explanation: Internal energy and transfers complete the account.

Quick check

The equation Δ(K+U)=Wother requires

Explanation: It is a rearrangement of the work–energy theorem.

Quick check

At terminal speed during a fall, net work over an interval is

Explanation: K is unchanged; drag and gravity work cancel.

Quick check

A 5% energy loss reduces v² by

Explanation: K is proportional to v² at fixed mass.

Quick check

On a rough approach before a spring, friction work should be counted

Explanation: All segments before the endpoint contribute to the energy balance.

Key Takeaways

Key Takeaways

• Mechanical energy may change while total energy remains accounted for. • Count every work term and avoid counting gravity or springs twice. • Check dissipative calculations by substituting the answer back into the energy balance.

Previous · Lesson 7

Spring Force and Elastic Potential Energy

Next · Lesson 9

Reading Potential-Energy Graphs