Work and Energy • Topic 1 of 3

Work and its Measurement

In everyday language we say a student doing homework or a person standing for hours holding a heavy bag is working hard. In physics, the word work has a much narrower, precise meaning. Work is said to be done only when a force acts on an object and the object moves in the direction of (or along the line of) that force. If there is no force, or no displacement, then no work is done, no matter how tired you feel.

Two conditions must both be satisfied for work to be done:

  • A force must act on the object.
  • The object must undergo a displacement (it must move).

When a constant force $F$ acts on an object and moves it through a displacement $s$ in the direction of the force, the work done is the product of the two:

$W=Fs$

Here $F$ is the force in newtons (N), $s$ is the displacement in metres (m), and $W$ is the work in joules (J). When the force and displacement are not in the same direction but make an angle $\theta$ with each other, only the component of force along the displacement does work, so $W=Fs\cos\theta$. For Class 9 we mostly use the simple case where force and motion are along the same line, i.e. $\theta=0$ and $\cos\theta=1$.

The SI unit of work is the joule (J), named after James Prescott Joule. One joule is the work done when a force of 1 newton moves an object through a distance of 1 metre in the direction of the force, so $1\,\text{J}=1\,\text{N}\times1\,\text{m}$. Lifting a 100 g apple (weight roughly 1 N) up by 1 m takes about 1 J of work.

Work is a scalar quantity — it has magnitude but no direction. However, its value can be positive, negative or zero depending on the angle between force and displacement:

  • Positive work — force has a component in the direction of motion ($\theta$ between $0^{\circ}$ and $90^{\circ}$). Example: pushing a box forward; gravity on a falling stone.
  • Negative work — force acts opposite to motion ($\theta$ between $90^{\circ}$ and $180^{\circ}$). Example: friction slowing a moving trolley; gravity on a ball thrown upward.
  • Zero work — either there is no displacement, or the force is perpendicular to the displacement ($\theta=90^{\circ}$). Example: a coolie carrying a load on his head and walking on level ground does no work against gravity, because the upward force and the horizontal motion are at right angles.

This is why simply pushing hard against a fixed wall — with great effort but zero displacement — counts as zero work in physics, even though your muscles tire.

Work done by a force moving a box through a displacementWork = Force x Displacement (W = Fs)BoxfinalForce FDisplacement sForce and motion in same direction: W = Fs (positive work)1 joule = 1 newton x 1 metre
1
Worked Example
A force of 15 N acts on a body and moves it through a distance of 4 m in the direction of the force. Calculate the work done.
Solution
  1. Given: force $F=15\,\text{N}$, displacement $s=4\,\text{m}$, force along motion.
  2. Work done $W=Fs$.
  3. $W=15\times4=60\,\text{J}$.

Answer: Work done $=60\,\text{J}$.

2
Worked Example
A boy pushes a wall with a force of 200 N for 5 minutes but the wall does not move. How much work does he do?
Solution
  1. For work to be done, displacement is required: $W=Fs$.
  2. The wall does not move, so displacement $s=0$.
  3. $W=200\times0=0\,\text{J}$.

Answer: Zero work is done, because there is no displacement.

3
Worked Example
A porter lifts a suitcase of mass 10 kg to a height of 1.5 m. Take $g=10\,\text{m/s}^2$. Find the work done against gravity.
Solution
  1. Force needed to lift = weight $=mg=10\times10=100\,\text{N}$.
  2. Displacement is upward, $s=1.5\,\text{m}$, in the direction of the applied force.
  3. $W=Fs=100\times1.5=150\,\text{J}$.

Answer: Work done $=150\,\text{J}$.

4
Worked Example
Friction of 8 N acts on a moving trolley as it travels 6 m. Calculate the work done by friction.
Solution
  1. Friction acts opposite to the motion, so the work is negative.
  2. Magnitude: $Fs=8\times6=48\,\text{J}$.
  3. Since the force opposes displacement, $W=-48\,\text{J}$.

Answer: Work done by friction $=-48\,\text{J}$ (negative work).

5
Worked Example
A satellite moves in a circular orbit around the Earth. The gravitational force on it points towards the centre. How much work does gravity do on the satellite in one revolution?
Solution
  1. At every instant the gravitational force is along the radius (towards the centre).
  2. The satellite's displacement is along the circle, i.e. perpendicular to the radius, so $\theta=90^{\circ}$.
  3. $W=Fs\cos90^{\circ}=Fs\times0=0$.

Answer: Zero work, because force is perpendicular to displacement.

6
Worked Example
A force of 50 N is applied at an angle of $60^{\circ}$ to the horizontal to drag a box 4 m along the floor. Find the work done. (Take $\cos60^{\circ}=0.5$.)
Solution
  1. Only the component of force along the displacement does work: $W=Fs\cos\theta$.
  2. Substitute: $W=50\times4\times\cos60^{\circ}$.
  3. $W=50\times4\times0.5=100\,\text{J}$.

Answer: Work done $=100\,\text{J}$.

Key Points

  • Work is done only when a force acts AND the object is displaced; $W=Fs$ when force is along the motion.
  • When force makes an angle $\theta$ with displacement, $W=Fs\cos\theta$.
  • The SI unit of work is the joule (J): $1\,\text{J}=1\,\text{N}\times1\,\text{m}$.
  • Work is a scalar but can be positive (force with motion), negative (force opposes motion) or zero.
  • Zero work occurs when displacement is zero, or when force is perpendicular to displacement ($\theta=90^{\circ}$).
Tap an option to check your answer0 / 4
Q1.Work is done by a force only if the object
Explanation: Work needs both a force and a displacement; without displacement $W=Fs=0$.
Q2.The SI unit of work is the
Explanation: $1\,\text{J}=1\,\text{N}\times1\,\text{m}$ is the SI unit of work.
Q3.A coolie carrying a load on his head walks on level ground. The work done against gravity is
Explanation: Gravity acts downward but motion is horizontal, so $\theta=90^{\circ}$ and $W=0$.
Q4.Friction acting on a moving body does
Explanation: Friction opposes motion ($\theta=180^{\circ}$), so the work it does is negative.