Work Formula: Definition, Formula, Derivation

Work Formula: Imagine sitting in your chair all day, in the same position, and studying nonstop. Imagine someone approaching you and saying, “Hey, you didn’t work all day!” What would your reaction be? You will, without a doubt, be enraged. But what if I told you that physics says you didn’t perform any work at all? But why is it the case? It is due to the fact that a job entails movement. Let’s find out more about work done, its formula, and how to calculate it.

What is Work?

Work is the external energy transferred from or to an object by applying a force in the direction of displacement of the object. Work is said to be done when a body displaces from its initial position under the action of an external force. Force and displacement are essential components for calculating the work done. Thus, it is important to realise that no work will be done unless the body gets displaced. So you can use all your energy in pushing a wall; unless the wall moves, your work done will be zero.

Here it is important to remember that work done on a body is accomplished not only by displacement of the body as a whole, but it can also be achieved by, say, compressing a gas or by rotating a shaft, and even by causing invisible motions of the particles within a body by an external magnetic force.

What is the Formula for Work Done?

Work is said to be done when the applied force displaces an object from its position. The work done by an object is equal to the dot product of force and the displacement vector. In simple words, Work done is equal to the product of the force acting on an object in the direction of displacement and the displacement of the object. Thus, the formula for work done can be given as:
\(W = \overrightarrow F .\overrightarrow s \)
\(W = \left( {F\,\cos \,\theta } \right)s\)

Where \(W\) is the work done
\(F\) is the force acting on the object.
\(s\) is the displacement.
\(\theta \) is the angle between the force and displacement vectors.
As we can see from above, the formula of work done involves the dot product of force and displacement. Force and displacement are both vector quantities, but we know that the scalar or dot product of two vector quantities results in a scalar quantity. Thus, Work done is a scalar quantity.

Work Formula Derivation

Work done by an external force is equal to the change in kinetic energy of the object. i.e. \(W = \Delta K\)
\(W = {K_f} – {K_t}\)

Let us assume that an object with final velocity \(v,\) initial velocity \(u,\) covers a distance \(d\) under a constant acceleration \(a.\)
\(\frac{1}{2}m\left[ {{v^2}} \right]\) represents the Final Kinetic Energy \(\left( {{K_f}} \right)\) of the object.
\(\frac{1}{2}m\left[ {{u^2}} \right]\) represents the Initial Kinetic Energy \(\left( {{K_t}} \right)\) of the object.
Thus, the work done can be given as:
\(W = \frac{1}{2}m{v^2} – \frac{1}{2}m{u^2}\)…..(i)
Using the equation of motion,
\({v^2} = {u^2} + 2\overrightarrow a .\overrightarrow d \)
\({v^2} – {u^2} = 2\overrightarrow a .\overrightarrow d \)
Substituting this in equation (i), we get:
\(W = \frac{1}{2}m \times 2\overrightarrow a .\overrightarrow d \)
\(W = m\overrightarrow a .\overrightarrow d \)

From Newton’s second law of motion, we know that force acting on an object, moving with constant acceleration, can be given as \(\overrightarrow F = m\overrightarrow a .\)
Thus, the expression for work done becomes,
\(W = \overrightarrow F .\overrightarrow d \)
The work done by a constant force is equal to the product of force and the displacement of the object in the direction of force, or the dot product of force vector and the displacement vector, i.e. \(W = \overrightarrow F .\overrightarrow d .\)

Work Formula Dimension and Unit

Work done depends on the scalar product between force and displacement vectors. The SI unit of force is newton \(\left( \rm{N} \right),\) and the SI unit of displacement is meter \(\left( \rm{m} \right).\) Thus, the SI unit of work can be calculated as:
\(W = F.s\)
SI unit: \(\left( {{\rm{newton}}} \right).\left( {{\rm{meter}}} \right) = \rm{N.m = J}\)
Thus, the SI unit of Work is Joule \(\left( \rm{J} \right).\)
The dimensions of force: \(F = \left[ {{M^{ – 1}}{L^1}{T^{ – 2}}} \right]\)
The dimensions of displacement: \(s = \left[ {{M^0}{L^1}{T^0}} \right]\)
Thus, the dimensions of work: \(W = \left[ {{M^1}{L^1}{T^{ – 2}}} \right].\left[ {{M^0}{L^1}{T^0}} \right] = \left[ {{M^1}{L^2}{T^{ – 2}}} \right]\)

Types of Work

Work is said to be done on an object when it experiences displacement. Work done can be of following three types:

a. Positive Work: When the force or component of force applied to an object causes it to displace in the direction of force or component of force, then the work done on the body is said to be positive. Thus, for positive work done, the angle \(\left( \theta \right)\) between force and displacement vectors must be \({0^{\rm{o}}} \le \theta < {90^{\rm{o}}}.\) For example, when a ball isly falling downwards, the force of gravity will act downwards, along with the ball’s motion. Thus, work done by gravity on the ball will be positive.

b. Negative Work: When the force (or component of force) applied to an object and displaced are in the opposite direction, then the work done by the force on the body is said to be negative. Thus, for negative work done, the angle \(\left( \theta \right)\) between force and displacement vectors must be \({\rm{90 < }}\theta \le {180^{\rm{o}}}.\) For example, when you throw a ball vertically upwards, the force of gravity acts downwards, opposite to the displacement of the ball. Thus work done by gravity will be negative.

c. Zero Work: When the force applied to the object does not cause any displacement in the object, the work done on the object is said to be zero. Also, no work is done on an object when the force acts perpendicular to the displacement. In this case, the angle between force and displacement vectors is \({90^{\rm{o}}}.\) For example, if you hold a heavy ball in your hand for hours. Since there is no ball displacement throughout, work done by your hand will be zero, even though your hand will feel tired over time.

The Area Under Force-Displacement Graph

The force acting on an object can be presented as a function of displacement graphically. Work done by a force can be calculated using the force vs displacement graph. Total work done will be equal to the area under the force vs displacement curve; here, keep in mind that areas above the position axis represent positive work and areas below the axis represent negative work.

Work Formula- Solved Examples

Q.1. A force of \(20\,\rm{N}\) is applied on a body which displaces it by \(2\,\rm{m}\) along its direction. Calculate the work done by using the Formula for Work.
Solution: We are given: Force \(\left( F \right) = 10\,\rm{N}\)
Displacement \(\left( d \right) = 2\,{\rm{m}}\)
Using Formula For Work, \(W = F.d\)
\(W = \left( {20\,\rm{N}} \right).\left( {2\,{\rm{m}}} \right)\)
\(W = 40\,\rm{J}.\)

Q.2. A coolie at a railway station carries a bag weighing \(200\,\rm{N}\) on his head through some distance. Calculate the work done by the force applied by the coolie to support this \(200\,\rm{N}\).

Solution: We are given: Weight of the bag \( = 200\,\rm{N}\)
In this case, the bag’s weight will act in the vertical direction, and coolie’s supporting force is vertically upwards. Also, he is moving in a horizontal direction. Hence, the displacement of the bag in the direction of the force (supporting force) is zero.
Using formula For work, \(W = F.d\)
\(W = \left( {200\,\rm{N}} \right).0\)
\(= 0\,\rm{J}\)
Thus, the work done by the supporting force applied by the coolie on the bag is zero.

Q.3. Calculate the amount of work done by a force of \(300\,\rm{N}\) in magnitude in moving an object through a distance of \(14\,\rm{m}\) if the object is horizontally dragged across the surface by this force.
Solution: We are given:
Force \( = 300\,\rm{N}\)
displacement \( = 14\,{\rm{m}}\)
Since force and displacement are in the same direction,
Thus, \(\theta = {0^{\rm{o}}}\)
\(W = F\,\cos \,\theta .d\)
\(W = 300 \times 14 \times \cos \,{0^{\rm{o}}}\)
\(W = 4200\,\rm{J}.\)

Summary

Work is defined as the energy transferred to or from an item by the application of force along a displacement in areas such as physics. It is frequently expressed as the product of force and displacement in its simplest form. If or when applied to an object, a force is said to produce positive work if it has a component in the direction of displacement of the site of application. When a force has a component that is in the opposite direction as the displacement at the point of application, it causes negative work.

We may say, for example, that when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball. A displacement is defined as a force multiplied by the distance to the ground.

Work is the external energy transferred from or to an object by applying a force in the direction of displacement of the object. Work is said to be done when a body displaces from its initial position under the action of an external force. The formula for work done can be given as:

\(W = \overrightarrow F .\overrightarrow s \) or \(W = \left( {F\,\cos \,\theta } \right)s.\)

Moreover, we learnt that the SI unit of work is Joule \(\left( \rm{J} \right).\) Additionally, depending upon the direction of force and displacement, work can either be positive, negative or zero.

Frequently Asked Questions (FAQs) on Work Formula

Q.1. What is work done?
Ans: Work is said to be done when an applied force causes an object to displace in its direction.

Q.2. What is the formula to calculate work done?
Ans: The formula to calculate work done is: \(W = \left( {F\,\cos \,\theta } \right)s\)

Q.3. Write the relation between work done and kinetic energy.
Ans: Work done is equal to the change in kinetic energy of the object.

Q.4. What is the SI unit of work done?
Ans: SI unit of work done is Joule.

Q.5. Can work done by an external force be zero? If yes, give an example.
Ans: Yes, work done by an external force will be zero when the force does not displace the object from its position. For example, when we try to push a brick wall, the wall does not move, although we feel drained after a while. Here, work done by us on the wall is zero.