In this post, you will learn how to calculate work done in physics. I will first explain the concept of work in physics and later solve a few problems that are to work done in physics.

## What is Work Done in Physics?

Work is the application of physical or mental effort to overcome an activity. When you pressed your television remote, work is involved. When you carry a bucket to fetch water, work is also involved. Knocking on your door requires energy to carry out the work (of knocking on the door). Whatever we do, we apply force to make the activity (work) a successful one.

Therefore, **we can define work done in physics as the product of force and the perpendicular distance in the direction where we apply the force.**

We use w as the symbol of work done in physics. Work is a scalar quantity, and we measure work in Joules (J).

The formula for calculating work done in physics is

**Work (w) = Force (f) x Distance (s)**

Which means

**w = f x s** [where w = work done in joules, f = force in newton, and s = distance in meters]

We can equally say that the formula for work done is

**w = ma x s** [because f = ma and m = mass of the object, while a = acceleration]

**The formula for work done along an inclined plane is w = f x s cosθ**

## Examples: How to Calculate Work Done in Physics

Here are a few examples that will guide you to understanding how to calculate work done in physics.

### Example 1

A drum of oil of a total mass of 100 kilograms rolls down from the floor of a lorry 3.0 meters high. Calculate the work done by gravity on the load.

**Solution**

**Data**

Mass of the drum, m = 100 kg

height = displacement = distance in a specified direction, s = 3 meters

w =?

and the formula for calculating work done is

w = f x s = ma x s

But we are dealing with gravitational force, and this implies that a = g

Thus,

w = mg x s

We can now plugin our data into the above formula

w = 100 x 10 x 3 = 3,000 Joules [Because g = 10ms^{-2}]

By converting the above answer to kilojoules, we will now have

w = 3 kJ [Because 1000 Joules = 1 kilojoules]

Therefore, the work done by the gravity on the load is 3 kilojoules.

### Example 2

With the help of a rope inclined at 30^{0} to the horizontal, a boy exerted a 25-newton pulling force at a desk and made it move a distance of 5 meters along the horizontal direction. Find the work done by the boy.

**Solution**

**Data:**

Always remember to extract your data from the question. We can see from the above question that

The angle made, θ = 30^{0}

The force, f = 25 Newton

Displacement, s = 5 meters

work done, w =?

Apply the formula for the work done in an inclined plane, which is w = f x s cosθ

Now, replace the above formula with the data you extracted to obtain

w = 25 x 5 x cos30^{0}

Therefore, w = 125 x 0.866025

Which implies that

w = 108.25 J

Therefore, the work done by the boy is 108 Joules

### Example 3

The mass of David Johns is 30 kilograms, and he climbed a flight of 20 steps each 15 centimeters high. How much work did David Johns do? [ take g = 10ms^{-2}]

**Solution **

**Data**

Mass of David Johns, m = 30 kg

Displacement covered, s = 20 steps x 15 cm = 300cm

We can now convert displacement into meters to suit our answer

Displacement, s = 300cm = (300/100) m = 3 meters

Work done =?

g = 10 ms^{-2}

By applying the formula for work done, w = mg x s

we will get

w = 30 x 10 x 3 = 900 J

Therefore, the work done by David Johns is 900 Joules.

### Example 4

A constant force of 40-newton acting on a body initially at rest gives it an acceleration of 0.1 ms^{-2} for 4 seconds. Calculate the work done by the force.

**Solution **

**Data**

Force, f = 40 N

Initial velocity, u = 0

acceleration, a = 0.1 ms^{-2}

time, t = 4 seconds

we can apply one of the equations of motion ( s = ut + 1/2 at^{2}) to find s

s = 0 x 4 + (1/2) x 0.1 x 4^{2}

Which becomes s = (1/2) x 0.1 x 16 = 0.8 meters

Now substitute our data above into the formula for work done ( w = f x s )

w = 40 x 0.8 = 32 J

**Therefore, the work done by the force is 32 Joules.**

### Example 5

A crane lifts a load of 1000 kilograms very slowly through a vertical distance of 150 centimeters. Calculate the work done against gravity if g = 10 ms^{-2}.

**Solution**

**Data**

Mass, m = 1000 kg

Distance, s = 150 cm = (150/100) m = 1.5 m

Work done, w =?

gravitational force, g = 10ms^{-2}

Plugin the above data into w = mg x s to get

w = 1000 x 10 x 1.5 = 15,000 Joules = 15 kJ

**Therefore, work done against gravity is 15 kilojoules**.

### Example 6

A loaded sack of total mass 100 kilograms falls down from the floor of a lorry 2.0 meters high. Calculate the work done by the gravity of the load.

**Solution**

**Data**

Mass, m = 100 kg

Distance, s (h) = 2 m

Work done, w =?

The formula for work done is

w = mg x s = mgh

Now, insert your data into the above formula

w = mgh = 100 x 10 x 2 = 2000 J = 2 kJ

**Therefore, the work done by gravity is 2 kJ **

### Example 7

Nazeef carries a 50-newton bag and climbs five steps of a staircase with a height of 5 meters. The bag was pulled away from his hands (due to its heaviness) with a force of 15 Newton at a constant speed of 0.1m/s through a horizontal distance of 25 meters. Find the work done by Nazeef for his entire motion.

**Solution**

**Data**

The force on the vertical axis, f_{v} = 50 N

Height along the vertical axis, s_{v} = 5 m

Angle on the vertical axis, θ_{v} = 0

Force on the horizontal axis, f_{h} = 15 N

Height along the horizontal axis, s_{h} = 25 m

The angle on the horizontal axis, θ_{h} = 0

The work done along the vertical axis, w_{v} = f_{v} x s_{v} cosθ

Hence

w_{v} = 50 x 5 x cos0 = 250 x 1 = 250 Joules

The work done along the vertical axis, w_{h} = f_{h} x s_{h} cosθ

Which implies that

w_{h} = 15 x 25 x cos0 = 375 x 1 = 375 Joules

**Therefore, the work done by Nazeef on his entire motion is **

w_{T} = w_{v} + w_{h} = 250 + 375 = **625 Joules**

*You may also like to read:*

What is the work done by the electric force to move a 1 c charge from a to b?

How to Calculate the Cost of Electricity Per kWh

Also, How to Calculate the Work Function of a Metal

**Reference**