Work Done: Positive, Negative, And Zero Work Explained

Work Done: Positive, Negative, and Zero Work Explained

Problem (Why students get stuck on “work done”)

You’ve solved dozens of force and motion problems, but “work done” still feels slippery. Why is pushing a wall for a minute “zero work,” while pulling a suitcase for two seconds is “positive work”? Why does braking a car do “negative work,” and what does that even mean physically? If the formula is just W = F * s * cos(theta), why do sign mistakes keep showing up in your answers?

Agitate (How misunderstandings create bigger problems)

Misreading the sign of work creates a chain reaction of errors:

• In mechanics, a wrong sign on work gives you the wrong change in kinetic energy. That means wrong speeds, wrong stopping distances, and wrong safety estimates.
• In real life, people confuse metabolic effort with mechanical work. You can get tired holding a heavy bag without moving it-your body spends energy-but the mechanical work on the bag is zero. Mixing those ideas leads to poor training plans in sports, and wrong expectations in jobs that require lifting and carrying.
• In exams, students forget to check the direction between force and displacement. They add magnitudes instead of adding signed works, so they miss the simple “net work equals change in kinetic energy” link.

Let’s fix all of that with a clean, step-by-step way to think about work.

Solution Overview (What you’ll learn and use)

In this lesson you will:

1. Pin down the definition of work in mechanics and the sign convention.
2. Classify work as positive, negative, or zero using an easy three-question test.
3. Compute work for constant and variable forces, including springs and friction.
4. Apply the work–energy theorem to check answers quickly.
5. Practice with realistic mini–case studies (braking a car, catching a cricket/baseball, pulling a suitcase, lifting and carrying in a warehouse).
6. Avoid common traps with a short checklist.

No fancy language-just clear reasoning, examples with numbers, and a method you can reuse.

1) What is “work done” in physics?

Definition: Work done by a force on an object is the product of the component of the force along the object’s displacement and the magnitude of that displacement.
For a constant force and straight-line motion:
W = F * s * cos(theta)

• F = force magnitude
• s = displacement magnitude
• theta = angle between the force vector and the displacement vector
• Unit: joule (J) = newton-meter (N·m)

Key idea: Work is a scalar (it can be positive, negative, or zero), but it is not a vector.

Who is doing work on whom? Always state it clearly: “work done by friction on the block,” “work done by the person on the box,” etc. Signs flip if you switch roles.

2) The sign of work: quick intuition

• Positive work: The force has a component in the same direction as the displacement. It adds energy to the object (increases its kinetic energy, or increases its potential energy if it’s a conservative force like gravity).
• Negative work: The force has a component opposite the displacement. It removes energy from the object (like braking).
• Zero work: Either there is no displacement, or the force is perpendicular to the displacement, or the force is zero.

Three-question test:

1. Is there displacement? If no, then W = 0.
2. If yes, is the force aligned with displacement? If mostly along, W > 0.
3. Is the force mostly opposite to displacement? If yes, W < 0.
4.  (If exactly perpendicular, W = 0.)

3) Conditions for non-zero work

To get non-zero work:

• The object must move (non-zero displacement).
• The force must have a non-zero component along the path of motion.

No displacement → no work, even if you sweat. Perpendicular force → no work, even if the force is large (like centripetal force in uniform circular motion).

4) Core examples with numbers

A) Positive work: Pulling a suitcase

You pull with a 50 N force on a handle that makes a 60° angle to the horizontal. The suitcase slides 10 m horizontally.
W = F * s * cos(theta) = 50 * 10 * cos(60°) = 500 * 0.5 = 250 J

Interpretation: Your pull added 250 J of energy to the suitcase (ignoring friction). If friction is present, some of that energy goes into thermal energy.

B) Negative work: Friction on a sliding box

A box slides 15 m on a rough floor. Kinetic friction is 20 N opposite the motion.

W_friction = -F_k * s = -20 * 15 = -300 J

Interpretation: Friction removed 300 J of mechanical energy from the box and converted it into heat.

C) Zero work: Centripetal force in uniform circular motion

A 1 kg mass moves at constant speed 3 m/s in a circle of radius 2 m. The centripetal force is

F_c = m * v^2 / r = 1 * 9 / 2 = 4.5 N

But centripetal force points toward the center, while instantaneous displacement is tangent to the circle-perpendicular. Therefore,
W_centripetal = 0

Interpretation: Speed (and kinetic energy) stay constant; the force only changes direction, not speed.

5) Variable forces: area under the curve

When the force changes with position, work equals the area under the F–s graph:

W = ∫ F(s) ds

Spring example

For a spring (Hooke’s law): F_spring = -k * x (negative sign shows force direction opposite displacement from equilibrium).

• Work done by the spring on the block as it goes from x = x_i to x = x_f:

W_by_spring = ∫ F_spring dx = ∫ (-k x) dx = -(1/2) k (x_f^2 - x_i^2)

• Work done on the spring by an external agent is the negative of that.

Concrete numbers: A spring with k = 200 N/m is compressed from x = 0 to x = 0.10 m. The external agent must do

W_on_spring = + (1/2) k x^2 = 0.5 * 200 * 0.10^2 = 1.0 J

This 1 J is stored as elastic potential energy. When released, the spring does +1 J of work on the block (ignoring losses).

6) The work–energy theorem

Statement: The net work done on an object equals its change in kinetic energy:

W_net = ΔK = (1/2) m v_f^2 - (1/2) m v_i^2

This is the fastest way to check signs:

• If the object speeds up, ΔK > 0 → W_net > 0.
• If it slows down, ΔK < 0 → W_net < 0.
• Constant speed? ΔK = 0 → W_net = 0 (though individual forces may do non-zero work that cancel).

7) Mini case studies with realistic numbers

Case 1: Braking a car (why negative work matters)

A 1000 kg car travels at 20 m/s on level road. The driver brakes with an average retarding force of 5000 N.
By work–energy:

W_net = ΔK = 0 - (1/2) m v^2 = - (1/2) * 1000 * 20^2 = -200,000 J

Assuming the braking force is the only horizontal force, its work is also the net work:

W_brake = -F * d  =>  -5000 * d = -200,000
=> d = 40 m

Takeaway: Negative work by brakes equals the car’s loss in kinetic energy. If the road is wet and the effective frictional force is smaller, the stopping distance increases. In practical driving, typical friction coefficients on dry asphalt are often around 0.7 (varies with tire and surface), which aligns with the idea that stronger friction (bigger opposing force) means shorter stopping distance.

Case 2: Catching a ball (force depends on stopping distance)

A 0.145 kg ball arrives at 30 m/s; you catch it by moving your hands back 0.20 m while stopping it.

Kinetic energy of the ball:

K_i = (1/2) m v^2 = 0.5 * 0.145 * 30^2 = 65.25 J

Your hands must do negative work of -65.25 J on the ball to bring it to rest. If the average stopping force is F_avg and stopping distance is d = 0.20 m:
W = -F_avg * d = -65.25  =>  F_avg = 65.25 / 0.20 = 326 N

Takeaway: Increasing the stopping distance reduces the required average force. That’s why players “give” with the ball, and why vehicles use crumple zones.

Case 3: Pulling a suitcase vs. carrying a bag

Pulling a suitcase horizontally at constant speed on wheels: Your horizontal pull does positive work against rolling resistance; friction in the wheels and floor likely does an equal magnitude of negative work. Net work is near zero if speed stays constant.

Carrying a bag at constant height at constant speed: The gravitational force is vertical; your displacement is horizontal. For gravity, theta = 90° → gravity does zero work. If friction is negligible and you move at constant speed on level ground, net work on the bag is roughly zero. Yet you still get tired because your muscles use metabolic energy for internal processes—not because the bag gains mechanical energy.

Case 4: Lifting in a warehouse

You lift a 15 kg box from the floor to a shelf 1.8 m high at constant speed.

• Work done by your upward force on the box (ignoring small accelerations):

W_you_on_box ≈ +m g h = +15 * 9.8 * 1.8 ≈ +265 J

• Work done by gravity on the box during the lift is -m g h ≈ -265 J.

If you then carry the box horizontally at constant height for 20 m on a smooth floor, gravity does zero work, and the normal force also does zero work (both perpendicular to displacement). If speed is constant and horizontal resistive forces are tiny, the net work on the box is approximately zero during the carry.

8) Multiple forces: add works, not just forces

If several forces act (pull, friction, gravity, normal), compute the work by each force and then add them:

W_net = W_pull + W_friction + W_gravity + W_normal + ...

• The normal force often does zero work when displacement is perfectly horizontal on a level floor.
• Gravity does positive work on a falling object (theta = 0°) and negative work on a rising object (theta = 180°).
• Friction typically does negative work relative to the direction of motion.

Check with work–energy: If your sum is positive, speed should increase (unless other forms of energy like potential are changing in your chosen system). If it doesn’t, re-check signs.

9) Conservative vs. non-conservative forces (why it matters)

• Conservative forces (gravity, ideal spring) have potential energies. The work they do depends only on initial and final positions, not the path.
• Non-conservative forces (friction, many drag forces) depend on the path; they usually convert mechanical energy into thermal energy.

Example: Raising a mass by height h increases gravitational potential energy by m g h, which equals the negative of gravity’s work (-m g h) and the positive work you must supply (at least +m g h, ignoring losses).

10) Typical sign patterns you can memorize

• Engine or person pulling/propelling in direction of motion: Positive work.
• Brakes, friction, air drag (opposing motion): Negative work.
• Gravity while object rises: Negative.
•  Gravity while object falls: Positive.
• Normal/reaction forces: Often zero work if motion is perpendicular to the force (e.g., sliding on a level frictionless surface).
• Centripetal force in uniform circular motion: Zero work.

11) Step-by-step method you can apply to any problem

1. Define the system and what force’s work you’re calculating (“work done by X on Y”).
2. Choose coordinates and draw a free-body diagram.
3. Mark the displacement direction of the object during the interval.
4. Find the angle theta between the force and displacement.
5. Compute work:
   For constant force along straight path: W = F * s * cos(theta).
   For variable force: use the area under F–s or W = ∫ F(s) ds.
6. Sum over forces if you need the net work.
7. Check with work–energy: W_net = ΔK. If your signs are right, the speed change will make sense.

12) More worked mini-problems (with answers)

Problem A: Force at an angle on rough floor

A 12 kg crate is pulled 8 m at constant speed by a rope at 30° above the horizontal with tension 40 N. The kinetic friction coefficient is 0.15. Find (i) work by the pull, (ii) work by friction, (iii) net work.

1. Normal force N = m g - T sin(30°) = 12*9.8 - 40*0.5 = 117.6 - 20 = 97.6 N.
2. Kinetic friction f_k = μ_k N ≈ 0.15 * 97.6 ≈ 14.64 N (opposes motion).

(i) Work by pull:

 W_pull = T * s * cos(30°) = 40 * 8 * 0.866 ≈ 277 J (positive).

(ii) Work by friction:

 W_fric = -f_k * s = -14.64 * 8 ≈ -117 J.

(iii) Net work:

W_net ≈ 277 - 117 = 160 J.

But the crate moves at constant speed; shouldn’t W_net = 0? What happened?
 We forgot one force: if speed is constant, there must be another negative work (like rolling resistance or a slightly larger friction than our rounded value). In reality, to keep constant speed, the horizontal component of pull equals total resistive force. With our rounded numbers, the mismatch (160 J over 8 m → 20 N) is the “missing” resistance. This is a useful reminder: use work–energy as a consistency check.

Problem B: Gravity on a rising elevator counterweight

A counterweight rises 5 m at constant speed. Work done by gravity on the counterweight:
W_gravity = -m g h (negative because displacement is upward). If m = 200 kg,
W_gravity = -200 * 9.8 * 5 = -9800 J.
Motor and rope forces must supply +9800 J (ignoring losses) to keep constant speed.

Problem C: Spring launcher

A 0.50 kg puck is launched by a spring (k = 250 N/m) compressed 0.12 m on a horizontal frictionless surface. Find the speed as it leaves the spring.
Energy view: spring does positive work W = (1/2) k x^2 = 0.5 * 250 * 0.12^2 = 1.8 J.
 By work–energy: ΔK = +1.8 J = (1/2) m v^2 → v = sqrt(2*1.8/0.50) = sqrt(7.2) ≈ 2.68 m/s.

13) Common misconceptions and how to avoid them

Misconception 1: “If I am tired, I must be doing mechanical work.”
 Not necessarily. Holding a heavy bag steady is zero mechanical work on the bag—no displacement—though your muscles still burn energy. Always separate mechanical work (force * displacement) from metabolic energy (body’s chemical energy use).

Misconception 2: “If a force acts, work must be done.”
 No. If displacement is zero (pushing on a rigid wall that doesn’t move), or if the force is perpendicular to motion (centripetal force), work is zero.

Misconception 3: “The sign of work depends on which direction I call positive.”
 Your coordinate choice doesn’t change whether the force is along or opposite the displacement. The sign of work follows cos(theta). If theta < 90°, work is positive; if theta > 90°, work is negative.

Misconception 4: “Net work is just sum of force magnitudes times distance.”
 No. You must add signed works from each force. A big positive and a big negative can cancel to zero, consistent with constant speed.

Misconception 5: “If kinetic energy doesn’t change, no force did work.”
 Not true. Different forces can do non-zero works that cancel. Example: pulling a crate at constant speed on rough ground—your pull does positive work; friction does an equal negative work.

14) A quick decision tree (mental flow)

Is there displacement?

• No → W = 0.
• Yes → continue.

Angle between force and displacement?

• < 90° → positive work.
• = 90° → zero work.
• > 90° → negative work.

Need net effect?

• Sum all works.
• Check with ΔK.

15) Short “real-world” checklist

• Driving: Braking does negative work equal to the loss in kinetic energy. Heavier vehicles or higher speeds require more energy removal, so stopping distances grow fast with speed (proportional to v^2 for fixed braking force).
• Sports: “Giving” with a catch increases stopping distance, reducing average force (same negative work spread over more distance).
• Lifting: Raising a load by height h requires at least m g h positive work on the load (ignoring losses). Carrying it horizontally at constant height doesn’t change its gravitational potential energy, and gravity does zero work during the carry.
• Circular motion: Forces pointing to the center do zero work if speed is constant.

16) Practice set (try on your own; answers given)

1. A 20 N horizontal pull moves a box 6 m on a frictionless surface. Work by the pull?
2.  Answer: W = 20 * 6 = 120 J (positive).
3. A 0.20 kg ball is thrown straight up at 12 m/s. Work done by gravity during the 5 m ascent?
4.  Answer: W_g = -m g h = -0.20 * 9.8 * 5 = -9.8 J.
5. A 60 N force acts at 45° to the horizontal and moves an object 4 m horizontally. Work by the force?
6.  Answer: W = 60 * 4 * cos(45°) ≈ 240 * 0.707 ≈ 170 J (positive).
7. A cyclist coasts to a stop due to air drag; W_drag = -1200 J. Change in kinetic energy?
8.  Answer: ΔK = -1200 J (by work–energy).
9. A 2 kg mass moves in a horizontal circle at constant speed. Work by the tension in the string over one full circle?
10.  Answer: 0 J (perpendicular to motion throughout).

17) Summary cheat sheet

Formula: W = F * s * cos(theta) (constant force, straight path).

Signs:

• Along displacement → W > 0
• Opposite displacement → W < 0
• Perpendicular or no displacement → W = 0

Variable forces: W = ∫ F(s) ds (area under F–s curve).

Work–energy theorem: W_net = ΔK.

Common zero-work forces: Normal on level surfaces (horizontal motion), centripetal force in uniform circular motion.

18) Final message (how to use this in exams and life)

When you see a work problem, don’t rush into numbers. First, identify who does work on what, then mark the displacement direction, and only then compute F * s * cos(theta). Sum signed works, and always finish with the work–energy check. This habit prevents sign mistakes, improves speed on exams, and sharpens your intuition for real tasks—from safe stopping in traffic to better catching techniques in sports and smarter lifting on the job.

If you want, I can turn this into a printable worksheet or add a diagram set (free-body sketches + F–s areas) to practice these steps.