Work

Work

In everyday language, the word 'work' often refers to any activity that requires physical or mental effort, or that takes up time and energy. We might say we did a lot of work studying for an exam, or that a labourer does hard work carrying bricks. However, in physics, the definition of 'work' is very specific and quantitative. It is a precise concept linked to force and displacement.

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Not Much ‘Work’ In Spite Of Working Hard!

Consider some common scenarios that we intuitively call 'hard work' but where physics might say otherwise:

Example 1: Pushing a heavy wall. If you push against a massive wall with all your might, you will feel tired and exert significant effort. Your muscles are working, expending energy, and you might sweat. However, if the wall does not move even an inch, there is no displacement. According to the physics definition, since there is no displacement caused by your force, the work done by your force on the wall is zero.

Example 2: Carrying a heavy load and standing still. Imagine holding a heavy suitcase steady on your head. You are exerting a significant upward force equal to the weight of the suitcase to prevent it from falling. You will feel the strain and might get tired. But if you remain standing in one spot, the suitcase is not moving. Since there is no displacement, the work done by the force you apply on the suitcase is zero.

Example 3: Carrying a heavy load and walking on a level road. Now, imagine walking horizontally on a flat road while carrying that heavy suitcase. You are still applying an upward force on the suitcase (against gravity). Your displacement is horizontal. The upward force you apply on the suitcase is perpendicular to the horizontal displacement. As we will see in the scientific definition, when the force and displacement are perpendicular, the work done by that force is zero.

These examples highlight the crucial difference between the everyday understanding of work (effort, exertion) and the physics definition, which requires both a force and a displacement caused by that force (or a component of that force) in the direction of motion.

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Scientific Conception Of Work

In physics, work is a mechanism of energy transfer. When a force acts on an object and causes it to move, work is done by that force on the object, and energy is transferred to or from the object.

For a constant force $\vec{F}$ acting on an object that undergoes a displacement $\vec{d}$, the work done $W$ by the force is formally defined as the scalar product (dot product) of the force vector and the displacement vector.

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Definition using Dot Product

The work done $W$ is given by:

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

The dot product of two vectors is a scalar quantity. This tells us that work is a scalar quantity; it has magnitude but no direction.

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Work as $Fd\cos\theta$

The dot product $\vec{F} \cdot \vec{d}$ can also be expressed in terms of the magnitudes of the vectors and the angle between them. If $F$ is the magnitude of the force $\vec{F}$, $d$ is the magnitude of the displacement $\vec{d}$, and $\theta$ is the angle between the direction of the force $\vec{F}$ and the direction of the displacement $\vec{d}$, then the work done is:

$ W = F d \cos\theta $

This formula shows that work done depends on three things:

1. The magnitude of the force ($F$).
2. The magnitude of the displacement ($d$).
3. The angle ($\theta$) between the force and the displacement.

Alternatively, we can think of $W = Fd\cos\theta$ as the product of the displacement $d$ and the component of the force parallel to the displacement ($F\cos\theta$). Or, equivalently, the product of the force $F$ and the component of the displacement parallel to the force ($d\cos\theta$). Only the component of force along the direction of displacement does work.

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Units of Work

The SI unit of work is the joule (pronounced 'jool'), abbreviated as J. It is named after the English physicist James Prescott Joule.

From the formula $W = Fd$, when $F$ is in Newtons (N) and $d$ is in metres (m), work is measured in Joule.

One joule (J) is defined as the amount of work done when a force of one newton (N) displaces an object by one metre (m) in the direction of the force.

$ 1 \text{ J} = 1 \text{ N} \cdot \text{m} $

Other units of work or energy include the erg (in the CGS system, $1 \text{ J} = 10^7 \text{ erg}$), and the electron volt (eV, often used in atomic and nuclear physics). A common commercial unit for electrical energy consumption in India is the kilowatt-hour (kWh). $1 \text{ kWh}$ is the energy consumed by a 1 kilowatt appliance running for 1 hour. Its conversion to joules is:

$ 1 \text{ kWh} = (1000 \text{ W}) \times (3600 \text{ s}) = 3,600,000 \text{ J} = 3.6 \times 10^6 \text{ J} $

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Work Done By A Constant Force

The sign of the work done depends on the angle $\theta$ between the force and the displacement. We can analyse three main cases:

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Positive Work

Work done is positive when the force (or its component) acts in the same direction as the displacement. This occurs when the angle $\theta$ between $\vec{F}$ and $\vec{d}$ is between $0^\circ$ and $90^\circ$ ($0^\circ \le \theta < 90^\circ$).

The maximum positive work is done when the force is exactly in the direction of displacement ($\theta = 0^\circ$). In this case, $\cos(0^\circ) = 1$, and $W = Fd$. This means the force is increasing the speed or energy of the object.

Examples of Positive Work:

• Pushing a cart forward (force and displacement are in the same direction).
• Work done by gravity when an object falls downwards (gravity and displacement are downwards).
• Work done by the engine's driving force on a car moving forward.

Answer:

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Negative Work

Work done is negative when the force (or its component) acts in the opposite direction to the displacement. This occurs when the angle $\theta$ between $\vec{F}$ and $\vec{d}$ is between $90^\circ$ and $180^\circ$ ($90^\circ < \theta \le 180^\circ$).

The maximum negative work is done when the force is exactly opposite to the direction of displacement ($\theta = 180^\circ$). In this case, $\cos(180^\circ) = -1$, and $W = -Fd$. Negative work is usually done by forces that oppose motion, causing the object to slow down or lose energy.

Examples of Negative Work:

• Work done by friction on a sliding object (friction opposes motion).
• Work done by air resistance on a moving object.
• Work done by gravity when an object is lifted upwards (gravity is downwards, displacement is upwards).

Answer:

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Zero Work

Work done is zero when the force acting on the object is perpendicular to the direction of displacement. This occurs when the angle $\theta$ between $\vec{F}$ and $\vec{d}$ is exactly $90^\circ$. In this case, $\cos(90^\circ) = 0$, and $W = Fd(0) = 0$.

It's important to note that for zero work to be done by a force, either the force or the displacement must be zero, or the force must be perpendicular to the displacement. If both force and displacement are non-zero, work is zero only if $\theta = 90^\circ$.

Examples of Zero Work:

• Work done by the gravitational force on an object moving horizontally (gravity is vertical, displacement is horizontal).
• Work done by the centripetal force on an object moving in a circular path (centripetal force is radial, displacement is tangential).
• As discussed earlier, carrying a load horizontally on a level road (force on load is vertical, displacement is horizontal).

Answer:

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Net Work Done

Often, multiple forces act on an object simultaneously. The total or net work done on the object can be calculated in two ways:

1. Calculate the work done by each individual force and then sum them up algebraically (taking into account positive and negative work).
2. Find the net force acting on the object ($\vec{F}_{net} = \sum \vec{F}_i$) and then calculate the work done by this net force over the displacement: $W_{net} = \vec{F}_{net} \cdot \vec{d}$.

Both methods yield the same result for constant forces.

The net work done on an object is related to the change in its kinetic energy by the Work-Energy Theorem (which will be discussed in detail in a later section): $W_{net} = \Delta KE = KE_{final} - KE_{initial}$.

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Work Done By A Variable Force

The formula $W = Fd\cos\theta$ is only valid when the force $\vec{F}$ is constant over the entire displacement $\vec{d}$. What happens if the force changes in magnitude or direction during the motion?

If the force is variable, we need to use integration to calculate the work done. We divide the total displacement into many small displacement elements $d\vec{r}$. Over each small element, the force can be considered approximately constant. The work done over this small displacement $d\vec{r}$ is $dW = \vec{F} \cdot d\vec{r}$. The total work done is the sum (integral) of all these small amounts of work over the entire path from initial point A to final point B:

$ W = \int_{A}^{B} \vec{F} \cdot d\vec{r} $

If the motion is along the x-axis and the force is $F_x(x)$ varying with position $x$, and the displacement is from $x_1$ to $x_2$, then $d\vec{r} = dx \, \hat{i}$ and $\vec{F} = F_x(x) \, \hat{i}$, assuming the force is in the direction of displacement $(\theta = 0^\circ)$. The work done is:

$ W = \int_{x_1}^{x_2} F_x(x) \, dx $

Graphically, if we plot the component of the force in the direction of displacement ($F_{||}$) versus the displacement ($d$), the work done is equal to the area under the curve between the initial and final positions.