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WORK POWER AND ENERGY

Work, Power, and Energy – Part 1: Work & Force Dynamics


1. Definition of Work

In physics, Work is defined as the scalar quantity that measures the energy transferred to or from an object when an external force causes a displacement of that object along the line of action of the force component.

Mathematical Definition: Work done ($W$) by a constant force is defined as the dot product (scalar product) of the force vector ($\vec{F}$) and displacement vector ($\vec{d}$):

$$W = \vec{F} \cdot \vec{d} = F d \cos\theta$$

Where:
• $F = |\vec{F}|$ is the magnitude of the applied force.
• $d = |\vec{d}|$ is the magnitude of the displacement.
• $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{d}$.


2. Physical Characteristics of Work

PropertySpecification / Value
Governing Formula$W = \vec{F} \cdot \vec{d} = F d \cos\theta$
SI UnitJoule (J) ($1\text{ J} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2/\text{s}^2$)
CGS UnitErg ($1\text{ erg} = 1\text{ dyne}\cdot\text{cm} = 1\text{ g}\cdot\text{cm}^2/\text{s}^2$)
Unit Conversion$$1\text{ Joule} = 10^7\text{ Ergs}$$
Dimensional Formula$[M^1 L^2 T^{-2}]$
Physical Quantity TypeScalar Quantity (It possesses magnitude only)

3. Work Done by a Constant Force & Case Analysis

A force is classified as a constant force when both its magnitude and direction remain unchanged throughout displacement.

Block Displacement (d) Force (F) θ

Figure 1: Representation of Force $\vec{F}$ acting at an angle $\theta$ relative to displacement $\vec{d}$.

Detailed Case Analysis based on Angle $\theta$:

CaseAngle ($\theta$)$\cos\theta$ ValueWork Done FormulaNature & Result
Case 1$\theta = 0^\circ$$\cos 0^\circ = +1$$W = +Fd$Maximum Positive Work
Case 2$0^\circ < \theta < 90^\circ$$\cos\theta > 0$$W = Fd\cos\theta$Positive Work
Case 3$\theta = 90^\circ$$\cos 90^\circ = 0$$W = 0$Zero Work
Case 4$90^\circ < \theta < 180^\circ$$\cos\theta < 0$$W = -Fd|\cos\theta|$Negative Work
Case 5$\theta = 180^\circ$$\cos 180^\circ = -1$$W = -Fd$Maximum Negative Work

4. Types of Work: Positive, Negative, and Zero Work

A. Positive Work ($W > 0$)

Work is positive when the force vector component acts in the same direction as displacement ($0^\circ \le \theta < 90^\circ$). Positive work increases energy.

  • Example 1: Work done by gravity on a body falling freely toward Earth ($\theta = 0^\circ$).
  • Example 2: Work done by a horse pulling a cart horizontally ($\theta = 0^\circ$).
  • Example 3: Work done by a stretching force applied to expand a spring.

B. Negative Work ($W < 0$)

Work is negative when the applied force component opposes the direction of displacement ($90^\circ < \theta \le 180^\circ$). Negative work removes energy.

  • Example 1: Work done by friction force on a sliding block ($\theta = 180^\circ$).
  • Example 2: Work done by gravity on an object thrown vertically upward ($\theta = 180^\circ$).
  • Example 3: Work done by resistive fluid drag on a sinking stone.

C. Zero Work ($W = 0$)

Work done is zero if force is zero ($F=0$), displacement is zero ($d=0$), or force is perpendicular to displacement ($\theta = 90^\circ$).

  • Condition 1 ($d = 0$): Pushing against a solid wall without moving it.
  • Condition 2 ($\theta = 90^\circ$): Work done by centripetal force on an orbiting satellite.
  • Condition 3 ($\theta = 90^\circ$): Carrying a heavy load on one’s head while walking horizontally across a flat platform.

5. Work Done by a Variable Force

When the force acting on an object continuously changes magnitude or direction over position, the standard formula $W = F d \cos\theta$ cannot be applied directly.

A. Mathematical Expression (Calculus Approach)

Summing all small work elements from initial position $\vec{r}_i$ to final position $\vec{r}_f$ gives:

$$W = \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r} = \int_{x_i}^{x_f} F_x \, dx + \int_{y_i}^{y_f} F_y \, dy + \int_{z_i}^{z_f} F_z \, dz$$

B. Graphical Interpretation ($F\text{-}x$ Graph Area)

The total work done by a variable force equals the area bounded by the Force-Position ($F\text{-}x$) curve and the position axis between limits $x_i$ and $x_f$.

Position (x) Force F(x) x_i x_f Work Done = Area

Figure 2: Area under the $F(x)$ vs. $x$ graph representing total work done.


6. Conservative and Non-Conservative Forces

A. Conservative Force

A force is conservative if the work done by or against it in moving a particle between two points depends solely on the initial and final positions, completely independent of the path taken.

Closed Loop Property: The total work done by a conservative force along any closed path is strictly zero:

$$\oint \vec{F}_{\text{cons}} \cdot d\vec{r} = 0$$

Examples: Gravitational force, Electrostatic force, Spring force, Magnetic force.

B. Non-Conservative Force

A force is non-conservative if the work done by or against it depends directly on the path taken during displacement.

Closed Loop Property: The total work done along a closed path is non-zero ($\oint \vec{F}_{\text{non-cons}} \cdot d\vec{r} \neq 0$).

Examples: Friction force, Viscous drag force, Air resistance.

Comparison: Conservative vs. Non-Conservative Forces


Section 7: Conceptual Questions & Solved Numerical Examples

Part A: Conceptual Understanding Questions

Q1. A coolie carries a heavy trunk on his head and walks along a horizontal platform. Is he doing any work against gravity?

Answer & Explanation:

No, work done against gravity is zero. The gravitational force acts vertically downwards, while the displacement is horizontal ($\theta = 90^\circ$). Since $W = Fd \cos(90^\circ) = 0$, no mechanical work is done against gravity.

Q2. Can the work done by a force ever be negative? Explain with a real-world example.

Answer & Explanation:

Yes. Work done is negative whenever the applied force opposes the direction of motion ($90^\circ < \theta \le 180^\circ$). For example, when brakes are applied to a moving car, kinetic friction acts opposite to displacement ($\theta = 180^\circ$), resulting in negative work.

Q3. Why is work done in a closed path zero for gravitational force but non-zero for friction force?

Answer & Explanation:

Gravitational force is a conservative force, meaning work depends only on initial and final positions ($\oint \vec{F} \cdot d\vec{r} = 0$). Friction is a non-conservative force that continuously opposes motion along every segment of the path, dissipating energy as heat ($\oint \vec{F} \cdot d\vec{r} \neq 0$).

Q4. What is the work done by the centripetal force on a satellite orbiting the Earth in a circular path?

Answer & Explanation:

Zero Joules. Centripetal force acts radially inward towards the center of orbit, while the instantaneous displacement vector is along the tangent ($\theta = 90^\circ$). Therefore, $W = Fd\cos(90^\circ) = 0$.

Q5. Is work a scalar or a vector quantity? How does it depend on vectors?

Answer & Explanation:

Work is a scalar quantity (it has magnitude but no direction). However, it is calculated from the dot product (scalar product) of two vector quantities: Force ($\vec{F}$) and Displacement ($\vec{d}$).

Q6. If a body is displaced perpendicular to the applied force, how much work is performed?

Answer & Explanation:

Zero work. When force and displacement are perpendicular ($\theta = 90^\circ$), $\cos(90^\circ) = 0$, making the net work done equal to zero regardless of the force magnitude.

Q7. How do you find the work done by a variable force from a Force-Displacement ($F-x$) graph?

Answer & Explanation:

The work done by a variable force is equal to the area under the Force vs. Displacement curve bounded by the initial ($x_1$) and final ($x_2$) displacement limits on the x-axis.

Q8. A person holds a heavy suitcase of $20\text{ kg}$ in their hand while standing still for 10 minutes. How much work is done?

Answer & Explanation:

Zero Joules. Although muscle fatigue occurs, mechanical work requires displacement ($d$). Since the displacement $d = 0$, $W = F \times 0 = 0$.

Q9. Can a non-conservative force store potential energy in a system?

Answer & Explanation:

No. Potential energy can only be defined for conservative forces ($F = -\frac{dU}{dx}$). Non-conservative forces (like friction or drag) dissipate energy irreversibly as heat or sound, so potential energy cannot be associated with them.

Q10. What is the relation between $1\text{ Joule}$ and $1\text{ Erg}$?

Answer & Explanation:

$1\text{ Joule}$ is the SI unit of work ($1\text{ N}\cdot\text{m}$), and $1\text{ Erg}$ is the CGS unit ($1\text{ dyne}\cdot\text{cm}$). Since $1\text{ N} = 10^5\text{ dynes}$ and $1\text{ m} = 10^2\text{ cm}$, the relation is: $1\text{ Joule} = 10^7\text{ Ergs}$.

Part B: Solved Numerical Examples

Problem 1: Work Done by Constant Force at an Angle

Statement: A force of $50\text{ N}$ acts on a block at an angle of $60^\circ$ to the horizontal. Calculate the work done in displacing the block horizontally by $4\text{ m}$.

Given: $F = 50\text{ N}$, $d = 4\text{ m}$, $\theta = 60^\circ$

Calculation: $W = Fd \cos\theta = 50 \times 4 \times \cos(60^\circ) = 200 \times 0.5 = 100\text{ J}$

• Answer: $W = 100\text{ Joules}$

Problem 2: Vector Dot Product Method

Statement: A force $\vec{F} = (3\hat{i} + 4\hat{j} – 2\hat{k})\text{ N}$ displaces an object from position $A(1, 2, 0)\text{ m}$ to $B(4, 6, 5)\text{ m}$. Find the work done.

Displacement Vector ($\vec{d}$): $(4-1)\hat{i} + (6-2)\hat{j} + (5-0)\hat{k} = (3\hat{i} + 4\hat{j} + 5\hat{k})\text{ m}$

Dot Product: $W = \vec{F} \cdot \vec{d} = (3\times3) + (4\times4) + (-2\times5) = 9 + 16 – 10 = 15\text{ J}$

• Answer: $W = 15\text{ Joules}$

Problem 3: Work Done by 1D Variable Force (Polynomial Integration)

Statement: A force $F(x) = (3x^2 + 2x)\text{ N}$ moves a particle from $x = 1\text{ m}$ to $x = 3\text{ m}$. Find the total work done.

Integration: $W = \int_{1}^{3} (3x^2 + 2x) dx = \left[ x^3 + x^2 \right]_{1}^{3}$

Limits Evaluation: $(3^3 + 3^2) – (1^3 + 1^2) = (27 + 9) – (1 + 1) = 36 – 2 = 34\text{ J}$

• Answer: $W = 34\text{ Joules}$

Problem 4: Work Done Against Gravity

Statement: A crane lifts a mass of $200\text{ kg}$ vertically upwards through a height of $15\text{ m}$ at constant speed. Calculate the work done against gravity ($g = 9.8\text{ m/s}^2$).

Formula: $W = mgh$

Calculation: $W = 200 \times 9.8 \times 15 = 29,400\text{ J} = 29.4\text{ kJ}$

• Answer: $W = 29.4\text{ kJ}$

Problem 5: Work Done by Kinetic Friction

Statement: A block of mass $5\text{ kg}$ slides on a rough horizontal surface with coefficient of kinetic friction $\mu_k = 0.2$. Find the work done by friction as the block slides $10\text{ m}$ ($g = 10\text{ m/s}^2$).

Friction Force ($f_k$): $f_k = \mu_k m g = 0.2 \times 5 \times 10 = 10\text{ N}$

Work Calculation: $W = f_k d \cos(180^\circ) = 10 \times 10 \times (-1) = -100\text{ J}$

• Answer: $W = -100\text{ Joules}$ (Negative work)

Problem 6: Finding Angle Between Force and Displacement

Statement: A force of $40\text{ N}$ displaces a body by $5\text{ m}$ performing $100\text{ J}$ of work. Determine the angle $\theta$ between force and displacement.

Formula: $\cos\theta = \frac{W}{F d}$

Calculation: $\cos\theta = \frac{100}{40 \times 5} = \frac{100}{200} = 0.5 \implies \theta = \arccos(0.5) = 60^\circ$

• Answer: $\theta = 60^\circ$

Problem 7: Work Done in Stretching a Spring

Statement: A spring has a spring constant $k = 200\text{ N/m}$. Calculate the work required to stretch it from its equilibrium length by $0.1\text{ m}$.

Formula: $W = \frac{1}{2} k x^2$

Calculation: $W = \frac{1}{2} \times 200 \times (0.1)^2 = 100 \times 0.01 = 1\text{ J}$

• Answer: $W = 1\text{ Joule}$

Problem 8: Work Done by a Spring Force (Variable Force)

Statement: A spring with $k = 400\text{ N/m}$ is already stretched by $0.1\text{ m}$. Calculate the additional work done to stretch it further to $0.2\text{ m}$.

Formula: $W = \frac{1}{2} k (x_2^2 – x_1^2)$

Calculation: $W = \frac{1}{2} \times 400 \times (0.2^2 – 0.1^2) = 200 \times (0.04 – 0.01) = 200 \times 0.03 = 6\text{ J}$

• Answer: $W = 6\text{ Joules}$

Problem 9: Area Under Force-Displacement Graph (Trapezoidal Region)

Statement: A force increases linearly from $0\text{ N}$ to $20\text{ N}$ over a displacement of $0$ to $4\text{ m}$, and then remains constant at $20\text{ N}$ from $4\text{ m}$ to $8\text{ m}$. Calculate total work done.

Triangular Area ($0$ to $4\text{ m}$): $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 20 = 40\text{ J}$

Rectangular Area ($4$ to $8\text{ m}$): $\text{base} \times \text{height} = (8-4) \times 20 = 4 \times 20 = 80\text{ J}$

Total Work: $W_{\text{total}} = 40 + 80 = 120\text{ J}$

• Answer: $W = 120\text{ Joules}$

Problem 10: Work Done in 3D Variable Force Field

Statement: A force field is given by $\vec{F} = (2x \hat{i} + 3y^2 \hat{j})\text{ N}$. Calculate the work done in moving a particle along a path from $(0,0)$ to $(2\text{ m}, 1\text{ m})$.

Integration Setup: $W = \int_{0}^{2} 2x \, dx + \int_{0}^{1} 3y^2 \, dy$

Evaluate $x$-part: $\left[ x^2 \right]_{0}^{2} = 2^2 – 0 = 4\text{ J}$

Evaluate $y$-part: $\left[ y^3 \right]_{0}^{1} = 1^3 – 0 = 1\text{ J}$

Total Work: $W = 4 + 1 = 5\text{ J}$

• Answer: $W = 5\text{ Joules}$


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FeatureConservative ForceNon-Conservative Force
Path DependencyIndependent of path takenDependent on path taken
Work in Closed LoopStrictly Zero ($\oint \vec{F} \cdot d\vec{r} = 0$)Non-Zero ($\oint \vec{F} \cdot d\vec{r} \neq 0$)
Energy PreservationTotal Mechanical Energy is conservedMechanical Energy is converted into heat/sound
Potential Energy FunctionCan be defined ($F = -\frac{dU}{dx}$)Cannot be defined