5.2 Apply the particle work–energy principle | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Work, Energy, and Power

Relating the work of forces to a particle’s change in speed

The particle work–energy principle is useful when the main unknown is a particle’s speed after moving between two positions. Instead of finding acceleration at every point, add the work done by the forces over the motion and relate that total to the change in kinetic energy. Work is a scalar, so this method can simplify problems even when the particle’s path is not a straight line. The key is to define the system and the two states clearly, then calculate each force’s work with the correct sign.

What you will learn

Define the particle system, observer, reference frame, and initial and final states for a work–energy problem.
Calculate the work of constant and position-dependent forces using force components or integration.
Apply the work–energy principle with consistent signs and SI units.
Check whether a result is physically reasonable using direction, dimensions, and limiting cases.

1. Define the system and the two states

In this lesson, the system is one particle of mass

m

. The observer measures its motion from a fixed reference frame, with positions and velocities described relative to that frame. Choose an initial state, labelled

1

, and a final state, labelled

2

. The particle’s displacement is the vector change in position; its path length is the distance travelled along the route. These are not generally the same: work depends on the route when a force varies with position or direction.

Kinetic energy depends on speed, not velocity direction. For a particle moving at speed

v

, its kinetic energy is

T=\frac{1}{2}mv^2

. The work of a force is positive when that force contributes energy along the motion and negative when it opposes the motion. A force perpendicular to the instantaneous motion does no work at that instant.

T=\frac{1}{2}mv^2

State the particle, observer, reference frame, and initial and final positions before calculating.
Position, displacement, path length, and speed describe different features of motion.
Kinetic energy is a scalar and cannot be negative.

2. Calculate work and use the principle

For a force of constant magnitude and direction, its work is the force component along the displacement multiplied by the displacement. If the angle between force and displacement is

\theta

, then

U=Fs\cos\theta

. Thus, a force in the direction of motion does positive work, a force opposite the motion does negative work, and a perpendicular force does zero work.

For a force that changes with position, add its small contributions along the path:

U_{1\to2}=\int_1^2 \mathbf{F}\cdot d\mathbf{r}

. Here,

d\mathbf{r}

is a small displacement along the path. In one dimension, this becomes the signed area under a force-versus-position graph. For several forces, calculate the work of each force and add the results.

The particle work–energy principle states that the sum of the work of all forces on the particle equals the change in kinetic energy. Include applied forces, gravity, normal forces, and friction whenever they act, but only forces that do work affect the sum. For example, a normal force perpendicular to a straight, level displacement has zero work. Do not assume a force does zero work without checking its direction against the displacement.

This principle does not require the net force to be constant. It follows from Newton’s second law and the definition of work, and it is especially convenient when forces and positions are known but acceleration as a function of time is not.

\sum U_{1\to2}=T_2-T_1

Work has units of joules, equivalent to newton-metres.
Use the signed component of force along the motion, not just the force magnitude.
Add the work of every force that acts over the interval.

3. A reliable problem-solving sequence

First define the particle system, observer, reference frame, initial state, and final state. Sketch the path or a free-body diagram if it helps identify the forces. A free-body diagram shows forces acting on the particle; it does not show the particle’s motion by itself.

Next choose coordinates and positive directions that make positions and force components clear. Record known values and the requested quantity. For a straight path, a coordinate along the path is often convenient. For a curved path, use the actual path when calculating position-dependent work.

Write one work contribution for each force, with a sign based on its direction along the motion, then use the work–energy equation. Solve symbolically before substituting values. If the result is a speed, take the nonnegative square root; a negative speed is not meaningful. If the equation predicts negative final kinetic energy, revisit the force work, signs, or the assumed motion.

Finally check units: work and kinetic energy must both be in joules. Check that the final speed is consistent with whether the net work is positive or negative. A useful limiting check is that zero net work gives unchanged speed. The principle determines speed between states, but by itself it does not specify the time taken.

v_2=\sqrt{v_1^2+\frac{2}{m}\sum U_{1\to2}}

A force–energy diagram or free-body sketch can help organize work terms.
Solve with signed work before inserting numbers.
Check the direction of each force’s work, dimensions, and limiting cases.

Worked example

A particle sliding down a rough incline

A particle of mass

4.0\,\mathrm{kg}

starts from rest and moves

3.0\,\mathrm{m}

down a straight incline at

30^\circ

below horizontal. The kinetic friction force has constant magnitude

5.0\,\mathrm{N}

and opposes the motion. Find the speed after the motion. Use

g=9.81\,\mathrm{m/s^2}

.

Forces on the particle

Schematic force directions for motion down a 30-degree incline; arrow lengths are not to scale.

Define the states and direction
The system is the particle, observed from a fixed ground frame. State 1 is at rest at the upper position; state 2 is 3.0 m farther down the incline. Take down the incline as the positive direction. The path is straight, so the component of gravity along it is $mg\sin 30^\circ$ .
Find the work of each force
Gravity acts partly along the displacement and does positive work. Friction acts against the displacement and does negative work. The normal force is perpendicular to the path and therefore does zero work.
$\sum U_{1\to2}=(mg\sin 30^\circ)s-f_k s$
Apply work–energy and solve
The initial kinetic energy is zero. Substituting the given values gives a net work of $43.9\,\mathrm{J}$ , which is positive, so the particle gains kinetic energy. Solving for the final speed gives approximately $4.69\,\mathrm{m/s}$ .
$v_2=\sqrt{\frac{2s}{m}(mg\sin 30^\circ-f_k)}=4.69\,\mathrm{m/s}$

Answer: The particle’s speed after moving down the incline is

4.69\,\mathrm{m/s}

.

Check: The gravity work is

58.9\,\mathrm{J}

and friction work is

-15.0\,\mathrm{J}

, giving

43.9\,\mathrm{J}

. This equals the final kinetic energy,

\frac{1}{2}(4.0)(4.69)^2\approx43.9\,\mathrm{J}

. The positive work and positive speed are consistent with motion down the incline.

Worked example

A variable horizontal force

A

2.0\,\mathrm{kg}

particle moves along a straight horizontal track from

x=0

to

x=4.0\,\mathrm{m}

. Its initial speed is

3.0\,\mathrm{m/s}

. A horizontal applied force varies with position as

F(x)=6.0+2.0x

newtons in the positive

x

direction. Find the final speed. Assume no other force does work.

Applied force along the track

The particle moves in the positive x direction; the applied force also points in that direction.

Define the system and states
The system is the particle in a fixed track-side frame. State 1 is at $x=0$ with speed $3.0\,\mathrm{m/s}$ ; state 2 is at $x=4.0\,\mathrm{m}$ . Choose positive $x$ to the right. The force varies along the path, so calculate its work by integration.
Integrate the force over the displacement
The force and displacement point in the same direction throughout, so the work is positive. Integrating from the initial position to the final position gives $40\,\mathrm{J}$ .
$U_{1\to2}=\int_0^4(6.0+2.0x)\,dx=40\,\mathrm{J}$
Find the final speed
The initial kinetic energy is $9.0\,\mathrm{J}$ . Adding the positive work gives a final kinetic energy of $49\,\mathrm{J}$ . Solve for the nonnegative speed to obtain $7.0\,\mathrm{m/s}$ .
$v_2=\sqrt{v_1^2+\frac{2U_{1\to2}}{m}}=7.0\,\mathrm{m/s}$

Answer: The final speed is

7.0\,\mathrm{m/s}

.

Check: The final kinetic energy is

\frac{1}{2}(2.0)(7.0)^2=49\,\mathrm{J}

, which is the initial

9.0\,\mathrm{J}

plus

40\,\mathrm{J}

of work. The force remains positive, so the increase in speed is consistent with the force direction.

Worked example

A force opposing motion

A

3.0\,\mathrm{kg}

particle moves along a straight horizontal path with initial speed

8.0\,\mathrm{m/s}

. A constant force of

12\,\mathrm{N}

acts opposite the direction of motion over a distance of

5.0\,\mathrm{m}

. Find the particle’s speed at the end of this distance.

Opposing force and motion

The force points opposite to the particle’s displacement.

Set the frame and direction
The system is the particle, observed from a fixed ground frame. State 1 is before the force acts over the specified distance; state 2 is 5.0 m farther along the initial direction of motion. Choose that direction as positive. The opposing force does negative work.
Calculate work and initial energy
The work is the force component along the displacement times the distance. Since the force points opposite to the displacement, its work is $-60\,\mathrm{J}$ . The initial kinetic energy is $96\,\mathrm{J}$ .
$U_{1\to2}=-Fs=-60\,\mathrm{J},\qquad T_1=\frac{1}{2}mv_1^2=96\,\mathrm{J}$
Solve for the final speed
Subtracting the work loss from the initial kinetic energy leaves $36\,\mathrm{J}$ of final kinetic energy. The resulting speed is $\sqrt{24}\,\mathrm{m/s}$ , or about $4.90\,\mathrm{m/s}$ . It remains real and positive because the opposing force has not removed all the initial kinetic energy.
$v_2=\sqrt{v_1^2-\frac{2Fs}{m}}=4.90\,\mathrm{m/s}$

Answer: The particle’s speed after 5.0 m is

4.90\,\mathrm{m/s}

.

Check: The final kinetic energy calculated from the reported speed is approximately

36\,\mathrm{J}

, matching

96\,\mathrm{J}-60\,\mathrm{J}

. If the opposing force acted over a sufficiently greater distance, the work equation would predict zero kinetic energy before the particle could continue; the stated distance is shorter than that stopping distance.

Common mistakes and how to avoid them

Using a force magnitude times distance without considering the force’s direction.

Correction: Use the signed force component along the displacement. Opposing forces do negative work.

Including the normal force’s magnitude as work automatically.

Correction: Work depends on the component along the displacement. A normal force perpendicular to the path does zero work.

Treating path length and displacement as interchangeable.

Correction: Use the actual path in the work calculation, especially when force varies with position or direction.

Using the work–energy equation with only one force when other forces also do work.

Correction: Sum the work of all forces acting on the particle between the selected states.

Reporting a negative final speed or accepting a negative final kinetic energy.

Correction: Speed is nonnegative. A negative predicted kinetic energy signals an inconsistent calculation or an unreachable assumed final state.

Lesson summary

Define the particle, frame, and initial and final states before calculating.
Work is the signed contribution of a force along the path; variable forces may require integration.
The sum of all force work equals the change in kinetic energy.
Check force directions, energy units, final speed, and whether the result is physically possible.

Check your understanding

Question 1

A force acts at right angles to a particle’s straight displacement. How much work does this force do over that displacement?

Zero
Positive work equal to force magnitude times distance
Negative work equal to force magnitude times distance
Work equal to the particle’s mass times distance

Show answer and explanation

Zero

The force has no component along the displacement, so its work is zero.

Question 2

A particle has initial kinetic energy

20\,\mathrm{J}

, and the total work on it is

-8\,\mathrm{J}

. What is its final kinetic energy?

$12\,\mathrm{J}$
$28\,\mathrm{J}$
$-12\,\mathrm{J}$
$8\,\mathrm{J}$

Show answer and explanation

12\,\mathrm{J}

The work–energy principle gives final kinetic energy as initial kinetic energy plus total work:

20\,\mathrm{J}-8\,\mathrm{J}=12\,\mathrm{J}

.

Question 3

A particle moves along the positive

x

direction while a constant force points in the negative

x

direction. What is the sign of the force’s work?

Negative
Positive
Zero in every case
It must equal the particle’s initial kinetic energy

Show answer and explanation

Negative

The force component along the displacement is negative, so the force does negative work.

Key terms

Work: The energy transferred by a force through displacement; its sign depends on the force component along the motion.
Kinetic energy: Energy associated with a particle’s motion, equal to one-half its mass times its speed squared.
Path: The route followed by a particle between two positions.
Work–energy principle: The statement that total work on a particle equals its change in kinetic energy.

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5.2 · Apply the particle work–energy principle

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Relating the work of forces to a particle’s change in speed

What you will learn

1. Define the system and the two states

2. Calculate work and use the principle

3. A reliable problem-solving sequence

A particle sliding down a rough incline

A variable horizontal force

A force opposing motion

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review