Work, Energy and Power

NCEES FE Civil spec: Dynamics, D. Work, energy, and power (e.g., particles, rigid bodies). FE Reference Handbook: Dynamics chapter, work, energy and power.

The work-energy method relates forces to changes in speed without solving for acceleration. It is often the fastest route on the FE Civil exam whenever a problem gives a distance and asks for a speed (or the reverse), and it is the only practical method for springs. Power is simply the rate of doing work.

What the NCEES spec covers

Section D, "Work, energy, and power", covers work of a force, kinetic and potential energy (gravity and springs), conservation of energy, the work-energy principle for particles and rigid bodies, power and efficiency.

Key concepts

Work is force times displacement in the direction of the force. Friction does negative work; normal forces on a stationary surface do none.
Work-energy principle. The total work of all forces equals the change in kinetic energy.
Conservation of energy. If only gravity and springs do work, kinetic plus potential energy stays constant. Friction converts mechanical energy to heat.
Rigid bodies carry rotational kinetic energy as well. A rolling wheel has both translational and rotational terms.
Power and efficiency. Output power is useful work per time. Input power is larger by the efficiency.

Core formulas

U_{1 \to 2} = \int F \cdot d r T_{1} + U_{1 \to 2} = T_{2}

T = \frac{1}{2} m v^{2} + \frac{1}{2} I_{G} ω^{2} V_{g} = m g h V_{e} = \frac{1}{2} k x^{2}

T_{1} + V_{1} = T_{2} + V_{2} (conservative forces only)

P = \frac{d U}{d t} = F v P = T ω η = \frac{P _{out}}{P _{in}}

Where it is in the FE Reference Handbook

The Dynamics chapter has sections on work and energy for particles and rigid bodies, including spring and gravity potential energy, and on power and efficiency. Unit conversions (hp, ft·lb/s, kW) are in the units and conversion factors pages at the front.

Worked example

A 2 kg block is dropped from rest 0.50 m above the top of a vertical spring with k = 2,000 N/m. Find the maximum compression of the spring and the peak spring force.

Step 1: energy balance. At maximum compression x the block is momentarily at rest, and it has fallen a total of 0.50 + x meters. Gravity work equals the spring energy stored:

m g (h + x) = \frac{1}{2} k x^{2} \Rightarrow 2 (9.81) (0.50 + x) = 1, 000 x^{2}

Step 2: solve the quadratic.

1, 000 x^{2} - 19.62 x - 9.81 = 0 \Rightarrow x = \frac{19.62 + 19.6 2 ^{2} + 4 ( 1 , 000 ) ( 9.81 )}{2 ( 1 , 000 )} = 0.109 m

Step 3: peak spring force.

F = k x = 2, 000 (0.1093) = 219 N

The peak force is about 11 times the block's weight (19.6 N), which is why impact loads matter. A related power check: a hoist lifting 800 kg at a steady 1.5 m/s needs $P = m g v = 11.8 kW$ of output, or 13.8 kW of input at 85% efficiency.

Common mistakes

Leaving out the extra drop distance x in the gravity term when a spring compresses.
Using $k x$ instead of $\frac{1}{2} k x^{2}$ for spring energy.
Forgetting rotational kinetic energy for rolling wheels or spinning drums.
Applying conservation of energy when friction acts. Include friction work as a loss.
Dividing by efficiency when you should multiply, or the reverse. Input is always larger than output.

See the work and power example.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

What is the kinetic energy of a 1,000 kg vehicle traveling at 20 m/s? See the worked solution or practice the Dynamics & Seismic deck
A 500 N applied force acts in the direction of motion of an object moving at a constant 3 m/s. What power does that force deliver? See the worked solution or practice the Dynamics & Seismic deck
A pump moves water at Q = 0.05 m³/s through a total head of 20 m with 80% pump efficiency. What is the required shaft input power? Use ρ = 1,000 kg/m³ and g = 9.81 m/s². See the worked solution or practice the Fluids & Hydraulics deck

Related Calculators and Examples

Frequently Asked Questions

When should I use work-energy instead of F = ma?

Use work-energy when the problem relates speeds to distances, or involves springs or variable forces along a path. Use F = ma when time or acceleration is asked for directly.

How is power related to force and speed?

Power equals force times velocity in the direction of the force, P = F v. For rotation, power equals torque times angular velocity in rad/s.

How does efficiency affect required power?

Efficiency is output power divided by input power, so the input power needed is the output power divided by the efficiency. A device that is 80 percent efficient needs 1.25 times the output power.

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