Work, Energy and Power • Topic 1 of 3

Work & the Work-Energy Theorem

Work in physics is not the same as everyday effort. A man holding a heavy suitcase while standing still gets tired, but in the language of physics he does zero work on the suitcase, because it has no displacement. Work is done only when a force succeeds in moving its point of application.

Work done by a constant force. If a constant force $\vec{F}$ acts on a body and the body undergoes a displacement $\vec{s}$, the work done is the scalar (dot) product $W=\vec{F}\cdot\vec{s}=Fs\cos\theta$, where $\theta$ is the angle between the force and the displacement. Work is a scalar; its SI unit is the joule (J), where $1\ \text{J}=1\ \text{N}\cdot\text{m}$. The sign of work depends entirely on $\cos\theta$:

  • Positive work ($0\le\theta<90^\circ$): the force has a component along the motion, e.g. gravity on a falling stone.
  • Zero work ($\theta=90^\circ$): the force is perpendicular to motion, e.g. centripetal force on a body in a circle, or the normal reaction on a block sliding horizontally.
  • Negative work ($90^\circ<\theta\le180^\circ$): the force opposes motion, e.g. friction, or gravity on a stone thrown upward.

Work done by a variable force. Most real forces change with position. Over a tiny displacement $dx$ the force is nearly constant, so the small work is $dW=F\,dx$. Adding all such bits gives $W=\int_{x_1}^{x_2} F\,dx$. On a force-displacement graph this integral is simply the area under the $F$-$x$ curve. This is the most powerful way to compute work for springs and other position-dependent forces.

Kinetic energy is the energy a body possesses because of its motion. For a body of mass $m$ moving with speed $v$, $KE=\frac{1}{2}mv^2$. It is a scalar, always positive, with the same unit (joule) as work. Since momentum $p=mv$, kinetic energy can also be written as $KE=\frac{p^2}{2m}$.

The work-energy theorem ties work and motion together: the net work done by all forces on a body equals the change in its kinetic energy, $W_{net}=\Delta KE=\frac{1}{2}mv^2-\frac{1}{2}mu^2$. A short derivation for constant acceleration: from $v^2=u^2+2as$ we get $\frac{1}{2}mv^2-\frac{1}{2}mu^2=mas=Fs=W_{net}$. The theorem holds for variable forces too and is often far quicker than using $F=ma$ directly, because it bypasses time and acceleration.

Force-displacement graph: shaded area under the curve equals work doneDisplacement xForce FArea = W = ∫ F dxx₁x₂F(x)
Solved Examples
1
Worked Example
A force of 50 N pulls a box through a displacement of 8 m. The force makes an angle of $60^\circ$ with the horizontal direction of motion. Find the work done.
Solution
  1. Step 1: Use $W=Fs\cos\theta$.
  2. Step 2: Substitute $F=50\ \text{N}$, $s=8\ \text{m}$, $\theta=60^\circ$, so $\cos 60^\circ=0.5$.
  3. Step 3: $W=50\times 8\times 0.5=200\ \text{J}$.

Answer: $W=200\ \text{J}$

2
Worked Example
A body of mass 2 kg is lifted vertically upward through 5 m at constant speed. Find (a) the work done by gravity and (b) the work done by the lifting force. Take $g=10\ \text{m/s}^2$.
Solution
  1. Step 1: Weight $=mg=2\times 10=20\ \text{N}$, directed downward; displacement is upward, so $\theta=180^\circ$ for gravity.
  2. Step 2: Work by gravity $=mgs\cos 180^\circ=20\times 5\times(-1)=-100\ \text{J}$.
  3. Step 3: At constant speed the lifting force equals the weight $=20\ \text{N}$ upward, along the motion, so work by lifting force $=20\times 5\times\cos 0^\circ=+100\ \text{J}$.

Answer: (a) $-100\ \text{J}$ by gravity; (b) $+100\ \text{J}$ by the lifting force.

3
Worked Example
A variable force acting along the x-axis is given by $F=(3x+4)\ \text{N}$, where x is in metres. Find the work done as the body moves from $x=0$ to $x=2\ \text{m}$.
Solution
  1. Step 1: For a variable force $W=\int_{x_1}^{x_2} F\,dx=\int_0^2 (3x+4)\,dx$.
  2. Step 2: Integrate: $W=\left[\frac{3x^2}{2}+4x\right]_0^2$.
  3. Step 3: Evaluate: $W=\left(\frac{3\times 4}{2}+8\right)-0=6+8=14\ \text{J}$.

Answer: $W=14\ \text{J}$

4
Worked Example
A car of mass 1000 kg accelerates from rest to 20 m/s. Using the work-energy theorem, find the net work done on the car.
Solution
  1. Step 1: Work-energy theorem: $W_{net}=\Delta KE=\frac{1}{2}mv^2-\frac{1}{2}mu^2$.
  2. Step 2: Here $u=0$, $v=20\ \text{m/s}$, $m=1000\ \text{kg}$, so $W_{net}=\frac{1}{2}\times 1000\times 20^2$.
  3. Step 3: $W_{net}=\frac{1}{2}\times 1000\times 400=200000\ \text{J}=2\times 10^5\ \text{J}$.

Answer: $W_{net}=2\times 10^5\ \text{J}$

5
Worked Example
A bullet of mass 20 g moving at 200 m/s strikes a wooden block and is stopped after penetrating 0.1 m. Find the average resistive force offered by the block.
Solution
  1. Step 1: By the work-energy theorem, work done by the resistive force $=\Delta KE=0-\frac{1}{2}mu^2$.
  2. Step 2: $m=0.02\ \text{kg}$, $u=200\ \text{m/s}$, so $\Delta KE=-\frac{1}{2}\times 0.02\times 200^2=-400\ \text{J}$.
  3. Step 3: The force opposes motion, so $W=-F\times s$, giving $-F\times 0.1=-400$, hence $F=\frac{400}{0.1}=4000\ \text{N}$.

Answer: $F=4000\ \text{N}$

6
Worked Example
A body moving in a horizontal circle of radius 3 m at constant speed experiences a centripetal force of 12 N. How much work does this force do in one complete revolution?
Solution
  1. Step 1: The centripetal force always points toward the centre, while the displacement is along the tangent, so the angle between them is $\theta=90^\circ$.
  2. Step 2: Work done $=Fs\cos 90^\circ$, and $\cos 90^\circ=0$.
  3. Step 3: Therefore the work done by the centripetal force is zero (the speed, and hence the kinetic energy, does not change).

Answer: $W=0\ \text{J}$

Key Points

  • Work by a constant force is $W=\vec{F}\cdot\vec{s}=Fs\cos\theta$; it is a scalar measured in joules.
  • Work is positive, zero or negative depending on whether $\theta$ is acute, $90^\circ$, or obtuse.
  • For a variable force, $W=\int F\,dx$, which equals the area under the force-displacement graph.
  • Kinetic energy of a moving body is $KE=\frac{1}{2}mv^2=\frac{p^2}{2m}$, always positive.
  • Work-energy theorem: net work done equals change in kinetic energy, $W_{net}=\Delta KE$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.A coolie carries a load on his head and walks on a level platform. The work done by him against gravity on the load is:
Explanation: The displacement is horizontal while gravity (and the supporting force) is vertical, so $\theta=90^\circ$ and $W=Fs\cos 90^\circ=0$.
Q2.The work done by a force $\vec{F}=(2\hat{i}+3\hat{j})\ \text{N}$ for a displacement $\vec{s}=(4\hat{i}+\hat{j})\ \text{m}$ is:
Explanation: $W=\vec{F}\cdot\vec{s}=(2)(4)+(3)(1)=8+3=11\ \text{J}$.
Q3.If the speed of a body is doubled, its kinetic energy becomes:
Explanation: Since $KE=\frac{1}{2}mv^2\propto v^2$, doubling $v$ multiplies $KE$ by $2^2=4$.
Q4.The area under a force-displacement graph represents:
Explanation: Work for a variable force is $W=\int F\,dx$, which is exactly the area under the $F$-$x$ curve.
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