In this article, we will look at how to calculate work done.** Work **is done when the point of application of a force moves along the force’s line of action. The **line of action** of the force is a line drawn through the point of application in the direction of the force. Suppose a constant force acts on an object and causes the object to move through a displacement in a direction parallel to its line of action, as shown below:

Then, the work done is given by,

Now consider the case where the force is acting at an angle to the displacement of the object. E.g., pushing a lawn mower—here, the lawn mower moves along the ground although the force is given at an angle to the ground. Here, only the component of force parallel to the ground is doing work. So to calculate work done in these cases, we multiply the *component of force in the direction of displacement by the displacement.*

That is,

This is, in fact, the definition of the scalar product between the force vector and the displacement vector. Therefore, we can also write work done as:

Note that work done is a scalar quantity. It has SI units of joules (J). 1 joule of work is done when the point of application of the force is displaced by 1 m, with the force having a component of 1 N in the direction of displacement.

If the force is not constant but rather a function of position , then the work done by the force to move the object from a position to another position is given by,

So if a force vs. distance graph is drawn, the work done to move the object from to is equal to the area under that graph between and .

As mentioned before, work done is not a vector quantity. However, we often take the work done by components forces acting in the direction opposite to the object’s displacement to be negative.

**Energy** is the ability to do work, and doing work transfers energy. The unit for measuring energy is also joules.

## How to Calculate Work Done – Examples

**Example 1**

*At a playground, a child sitting in a toy cart is pulled forward by her friend, who pulls the cart with a force of 60 N along a rope attached to the cart. The rope makes an angle of 35 ^{o} to the ground. Calculate work done by the child’s friend to pull the child forward by 20 m.*

We have , and .

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**Example 2**

*The engine of a car of mass 1500 kg suddenly stops working. The car travels 160 m before coming to a complete stop after 13 s. Assuming the only force responsible for stopping the car is the friction between car’s tyres and the road, find how much work needed to be done by this force to bring the car to a complete stop.*

For the car, we have , , , .

To find the acceleration, we use (which is an altered form of the more familiar ). Then,

.

The size of average friction is then,

.

Since friction acts through a distance of 160 m, work done is:

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**Example 3**

**The force of gravity between the Earth and the Sun is 3.5×10 ^{22} N. If the distance between the Sun and the Earth is 1.59×10^{11} m, calculate work done by the Sun to keep the Earth in its orbit for one full year. Assume the orbit of Earth around the Sun is circular.**

This is a trick question! It would *seem* that, to find the work done, you first find the total length of Earth’s orbit using , and then multiply that by the force. However, the Earth is in uniform circular motion around the Sun. Therefore, the force from the Sun is centripetal. i.e. it is always perpendicular to the Earth’s direction of motion. Consequently, at any given time the Sun has *no* force component along the direction of Earth’s motion. So the Sun does no work to keep the Earth in orbit. That is, the work done is 0. In general, no work needs to be done to keep an object in uniform circular motion.