Work Done by a Constant Force

A second way to look at motion

Newton's laws describe motion with vectors: forces, accelerations and velocities. For a body pushed by several forces along a curved path, the vector equations become long. Physics has a second way to study motion, built on a scalar quantity called energy. Energy can pass from one body to another and can change its form, but it is never lost. A force moves energy into a body or out of it, and this transfer is called work.

Work along the line of the force

Push a trolley with a steady force \(F\) along a straight corridor. The point where your hands press on the trolley is the point of application of the force. Let it move a distance \(d\) in the direction of the push. The work done by your push on the trolley is \(W = Fd\). Twice the force over the same distance is twice the work. The same force over twice the distance is also twice the work.

A force at an angle

Often the force and the displacement point in different directions. A child pulls a toy cart by a string that slopes upward at an angle \(\theta\) to the floor. The cart moves along the floor. Split the pull into two components. The component \(F\cos\theta\) lies along the displacement. The component \(F\sin\theta\) is at right angles to it. Only the first component drives the cart along. The second only lifts a little on the cart, which goes nowhere upward. So the work is

\[ W = (F\cos\theta)\,d = Fd\cos\theta \]

Here \(\theta\) is the angle between the force and the displacement, not the angle with the floor or with the vertical.

Work is a scalar product

The expression \(Fd\cos\theta\) is the scalar product of the two vectors, written \(\vec F\cdot\vec d\). When the vectors are given in components, the scalar product is the sum of the products of matching components: \(\vec F\cdot\vec d = F_x d_x + F_y d_y + F_z d_z\). The answer is a single number with a sign but no direction. So work is a scalar, although it is built from two vectors.

The same product can be read the other way round. \(Fd\cos\theta\) is also \(F \times (d\cos\theta)\): the force times the component of the displacement along the force. Use whichever reading is easier. For gravity, which always points straight down, the second reading is the natural one, as Lesson 3 will show.

The joule

Force is measured in newtons and distance in metres, so work is measured in newton metres. This unit is called the joule, written J. One joule is the work done when a force of one newton moves its point of application one metre along its own direction. In base units, \(1\,\text{J} = 1\,\text{N m} = 1\,\text{kg m}^2/\text{s}^2\).

How big is a joule?

A joule is small. An apple of mass \(100\,\text{g}\) weighs about \(1\,\text{N}\), so lifting it by one metre takes about \(1\,\text{J}\). A student of mass \(50\,\text{kg}\) who climbs one floor of a building, \(3\,\text{m}\) high, does \(50 \times 10 \times 3 = 1500\,\text{J}\) of work against gravity. Work on people, vehicles and lifts runs to thousands and millions of joules. So if a calculation says that carrying a person up a staircase takes \(4\,\text{J}\), the arithmetic has gone wrong somewhere.

The sign of work comes from the angle

Look at the factor \(\cos\theta\) in \(W = Fd\cos\theta\). The sizes \(F\) and \(d\) are never negative. So the sign of the work comes from the angle alone. Below \(90°\), \(\cos\theta\) is positive and the work is positive: the force puts energy into the body. Above \(90°\), \(\cos\theta\) is negative and so is the work: the force takes energy out. At exactly \(90°\), \(\cos\theta = 0\) and the work is zero.

Move the slider from \(0°\) to \(180°\). At \(0°\) the whole \(50\,\text{N}\) lies along the displacement, and the work is \(50 \times 4 = 200\,\text{J}\). At \(37°\) the component along the motion is \(40\,\text{N}\) and the work is \(160\,\text{J}\). At \(90°\) the component vanishes, and so does the rectangle. Beyond \(90°\) the component points backward, the rectangle drops below the axis, and the work is negative. At \(180°\) it reaches \(-200\,\text{J}\).

Everyday forces show all three signs. The pull of an engine on a train that speeds up does positive work. Gravity on a ball thrown straight up does negative work while the ball rises, because the ball moves up and gravity pulls down. Kinetic friction on a box sliding across a floor does negative work. The normal force of the floor on that sliding box does no work at all.

Two different ways to get zero

Zero work happens in two separate ways. In the first, the point of application does not move, so \(d = 0\). You push hard on a wall, and the wall does not move; your push does no work on the wall. A weightlifter holding a barbell still above his head does no work on it. In the second way, the body moves, but the force is at right angles to the displacement, and \(\cos 90° = 0\). The upward push of the porter's head on the trunk is at right angles to the trunk's level walk. This second kind of zero holds however far the body goes.

Now return to the porter. Lifting the trunk from the platform to his head, he pushes upward and the trunk moves upward. That work is positive and large: for a \(30\,\text{kg}\) trunk lifted \(1.6\,\text{m}\), it is \(30 \times 10 \times 1.6 = 480\,\text{J}\). Walking along the level platform, his head pushes the trunk upward while the trunk moves horizontally. The angle is \(90°\), so that push does no work at all. His tiredness on the walk is real, but it is energy spent inside his own body, not energy given to the trunk.