Scivra

Preview

Work-Energy Theorem

Net work equals change in kinetic energy

Push a filing cabinet across an office floor and you already know the punchline of this lesson in your shoulders: the harder you push and the farther it slides, the more energy you pour in — but the floor takes its cut. In this lab a green push-bar drives a block down a 12-metre track while a live ledger keeps the books. Every metre, the W_applied row grows by exactly the force you dialed in; the W_friction row siphons off μmg per metre as heat; and the amber W_net row — the difference between the two — lands, to the decimal, on the block's kinetic energy. That identity is the work-energy theorem: net work is not roughly, not usually, but exactly the change in KE, even while friction is draining energy out of the motion. You will watch the ledger balance on a rough floor, then on a frictionless one where nothing is siphoned, and finally meet the case that breaks intuition the other way: a 50 N shove on a heavy, sticky crate that goes nowhere at all — maximum effort, zero work, because the block never moves a millimetre. By the end, 'force' and 'work' will be two different words in your head, and the ledger will be why.

What you'll be able to do

  • Calculate the work done by each constant force on a block on a horizontal surface — W_applied = F_app·d positive, W_friction = μmg·d negative — and explain why the normal force and gravity do zero work on level ground
  • Apply the work-energy theorem W_net = ΔKE to predict the block's speed and kinetic energy after a known displacement, summing signed work from all forces rather than crediting the applied force alone
  • Predict the breakaway condition: the block only accelerates when F_app > μmg, and explain friction's negative work as a transfer of kinetic energy into thermal energy rather than a disappearance of energy

Formulas

Wnet=ΔKE=12mv2−12mv02W_{net} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2
Work-Energy Theorem: net work equals change in kinetic energy
W=F⋅d⋅cos⁡θW = F \cdot d \cdot \cos\theta
Work done by a constant force over displacement
P=Wt=F⋅vP = \frac{W}{t} = F \cdot v
Power: rate of doing work or force times velocity
ΔEmech=Wfriction=−μmg⋅d\Delta E_{mech} = W_{friction} = -\mu mg \cdot d
Mechanical energy lost to kinetic friction on the level track

Make a prediction

Default Rough Surface setup: a 20 N applied force pushes a 5 kg block (μ = 0.30, g = 9.8 m/s²) down the full 12 m track. Before you press Play — what does the W_net = ΔKE row show at the end of the run, and how does it compare with the W_applied row?

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Your prediction

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Answer: W_net ≈ 63.6 J — well under the 240 J of applied work, because friction (14.7 N) drains 176.4 J over the same 12 m and only the difference lands in kinetic energy

First check the block moves at all: kinetic friction is f_k = μmg = 0.30 × 5 × 9.8 = 14.7 N, which is less than the 20 N push, so the engine releases the block with a net force of 20 − 14.7 = 5.3 N (acceleration 5.3/5 = 1.06 m/s²). Over the 12 m track the ledger accumulates W_applied = 20 × 12 = 240 J and W_friction = 14.7 × 12 = 176.4 J, so W_net = 240 − 176.4 = 63.6 J. The work-energy theorem says this must equal ΔKE: ½ × 5 × v² = 63.6 J gives v ≈ 5.04 m/s at the end of the run — exactly what the closed form v = √(2ad) = √(2 × 1.06 × 12) predicts. Option B is the targeted misconception: it credits the block with the applied force's entire work and forgets that W_net sums ALL forces — friction's negative work is not optional. Option C overcorrects: friction opposing motion does not pin the block — only friction ≥ the applied force (the Heavy Load case, 73.5 N vs 50 N) can do that. Here 14.7 N < 20 N, so the block accelerates and banks 63.6 J of KE.

Quiz (0/3)

The Frictionless preset pushes a 5 kg block from rest with 25 N over the full 12 m track. What is its speed at the end?

On the default Rough Surface run (μ = 0.30, m = 5 kg, d = 12 m), how much energy is lost to friction?

A 5 kg block reaches 6 m/s after 10 seconds of pushing. What is the average power delivered by the net force?

You can now

  • Calculate the work done by each constant force on a block on a horizontal surface — W_applied = F_app·d positive, W_friction = μmg·d negative — and explain why the normal force and gravity do zero work on level ground
  • Apply the work-energy theorem W_net = ΔKE to predict the block's speed and kinetic energy after a known displacement, summing signed work from all forces rather than crediting the applied force alone
  • Predict the breakaway condition: the block only accelerates when F_app > μmg, and explain friction's negative work as a transfer of kinetic energy into thermal energy rather than a disappearance of energy

What you'll learn

  • What Is Work?. Work is energy transferred by a force acting over a displacement. Only the component of force parallel to the displacement does work. A force perpendicular to motion (like the normal force on a flat surface) does zero work, no matter how large it is.
  • Kinetic Energy. Kinetic energy is the energy of motion: ½mv². It depends on both mass and the square of velocity — so doubling speed quadruples kinetic energy. This is why car crashes at high speed are so much more devastating than at low speed.
  • The Work-Energy Theorem. The net work done on an object equals its change in kinetic energy. This powerful theorem connects force-based analysis (Newton's laws) with energy-based analysis. It works even when multiple forces act simultaneously — just sum all the work contributions.
  • Conservative vs Non-Conservative Forces. Gravity is conservative — work depends only on height change, not the path taken. Friction is non-conservative — it always removes mechanical energy as heat. On a rough floor, the block banks less kinetic energy than the push alone would provide because friction steals part of every metre's work as thermal energy.

Step-by-step

  1. Set the applied force, block mass, and friction coefficient μ (Pro), then press Play — the force pushes the block across a horizontal surface while friction drags against it.
  2. Watch the KE bar fill as the readouts track velocity, acceleration, normal force, and displacement.
  3. The work ledger shows W_applied (input), W_friction (lost), and W_net — verify that W_net = ΔKE at every snapshot.
  4. Try the presets: Rough Surface, Frictionless, and Heavy Load.

Key formulas

  • Wnet=ΔKE=12mv2−12mv02W_{net} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2Work-Energy Theorem: net work equals change in kinetic energy
  • W=F⋅d⋅cos⁡θW = F \cdot d \cdot \cos\thetaWork done by a constant force over displacement
  • P=Wt=F⋅vP = \frac{W}{t} = F \cdot vPower: rate of doing work or force times velocity
  • ΔEmech=Wfriction=−μmg⋅d\Delta E_{mech} = W_{friction} = -\mu mg \cdot dMechanical energy lost to kinetic friction on the level track

Frequently asked questions

The Frictionless preset pushes a 5 kg block from rest with 25 N over the full 12 m track. What is its speed at the end?
The correct answer is: v ≈ 11.0 m/s. No friction, so W_net = W_applied = 25 × 12 = 300 J. Then ½mv² = 300 J → v = √(2 × 300 / 5).
On the default Rough Surface run (μ = 0.30, m = 5 kg, d = 12 m), how much energy is lost to friction?
The correct answer is: W_f ≈ 176 J. F_k = μmg = 0.30 × 5 × 9.8 = 14.7 N. W_friction = f_k × d = 14.7 × 12 — the ledger's red row at the end of the run.
A 5 kg block reaches 6 m/s after 10 seconds of pushing. What is the average power delivered by the net force?
The correct answer is: P = 9 W. W_net = ΔKE = ½mv². P_avg = W_net / t.