We may understand work as in doing something or spending effort for our daily living. But, work in physics does not simply mean like that. Physics has a precise, accurate, specific and scientific definition of work and sometimes this definition does not agree with our everyday use of the word work. This page will help us understand more about work in terms of Physics.

There are two important components of work to be considered: force and displacement. Scientifically, work is done on an object when the force acted on the object causes the object to move. The general equation for work is: \({W} = F_{net} \cos\theta \cdot d\)


\(W=\text{work (in Joules)}\)

\(F=\text{magnitude of the force (in Newtons)}\)

\(d=\text{magnitude of the displacement (in meters)}\)

Consider the following situations:

  1. Pushing a box against the floor.
  2. Lifting a box from the floor.
  3. Keeping the box at shoulder height.

Picture 1 shows a boy pushing a box to the left. Since the direction of the force is going to the left, the box also moves towards left. In this case, the force is parallel to the displacement thus the angle is zero. Mathematically, 

\(W_{boy}=F_{boy}\cos\theta \cdot d =F_{boy}\cos 0 \cdot d\)

since \(\cos 0=1\), the equation can be simplified as

\(W_{boy}=F_{boy}\cdot d\)

As the box moves against the floor, another force is acting on the box – friction. Frictional force is acting on the box in an opposite direction. Thus, the angle is 180º. The negative work means that friction causes the box to slow down.

\(W_{friction}=F_{friction}\cos\theta \cdot d =F_{boy}\cos 180^\circ \cdot d\)

since \(\cos 180^\circ=-1\), then we can simplify the equation as

\(W_{friction}=-F_{friction} \cdot d\)

To solve for the total work done on the box, we just simply add the work done by the man and work done by friction. Therefore,

\(W_{total}=W_{boy} + W_{friction}=F_{boy}\cdot d + (-F_{friction}\cdot d)=(F_{boy}-F_{friction})d\) 

Picture 2 shows a man lifting a box from the floor. When a box is lifted upward, the direction of the box is also upward giving work as

\(W_{man}=F_{man}\cdot d \)

Is there another force acting opposite on the box? Yes, that is gravitational force which acts on the box in a downward direction, thus work done by gravity is

\(W_{gravity}=-F_{gravity}\cdot d\)

The total work done on the box is

\(W_{total}=W_{man} + W_{gravity}=F_{man}\cdot d + (-F_{gravity}\cdot d)=(F_{man}-F_{gravity})d\)

What if you kept the box at shoulder level? Is there work done on the box? Even if the box is kept at shoulder level for a long time, there is no work done on the box because there is no displacement. In equation form,

\(W_{total}=W_{man} + W_{gravity}=F_{man}\cdot d + (-F_{gravity}\cdot d)=(F_{man}-F_{gravity})0=0\)

In conclusion, work is done on an object when a force acting on an object causes the object to move in the direction of the force. Even if someone may tire holding, pushing, or lifting something as long as there is no displacement of the object, the work done is zero.

The following examples will help us understand more about work.

1. How much work is done by Alex in pushing his car for a distance of 150 meters with a force of 90 N?

\(\; \; \; \; \; \quad \quad \text{Given:}\\ \; \; \; \; \; \quad \quad \quad F=90\; N \\ \; \; \; \; \; \quad \quad \quad d=150\; m\)

To solve the problem, we can use the simplified form of the formula for work, that is, \(W=F\cdot d.\)

Substituting the given values of force and displacement, we have

\(W=F\cdot d\\ \, \quad = 90\; N \cdot 150\; m\\ \, \quad = 13, 500 \;J \)

Therefore, the work done by Alex in pushing his car is 13, 500 Joules.

2. The construction worker uses a rope to pull the 50-kg block from the ground to the second floor of the building. If the block is lifted 10 meters from the ground, what is the work done by the worker?

\(\; \; \; \; \; \quad \quad \text{Given:}\\ \; \; \; \; \; \; \quad \quad \quad m=50 \;kg\\ \; \; \; \; \; \; \quad \quad \quad d=10 \;m\)

The free-body diagram at the right will help us solve the problem. Assuming that the force needed to lift the blocks is equal to the force of gravity on the object. Thus, we can get the value of the force using the formula for weight or gravitational force which is \(Weight = F = mg\), then

\(F=mg\\ \; \; \,=(50\; kg)(9.8\; N/kg)\\ \; \; \, =490\; N\)

Use this value to solve for work, we have     

\(W=F\cdot d\\ \; \; \;\,= 490\;N \cdot 10\; m\\ \; \; \; \, = 4,900 \ J\)

Thus, the worker does 4, 900 J of work in lifting the blocks from the ground.

3. A crate with a mass of 20-kg is being pulled by a force of 40 N at an angle of 45º to the horizontal. If the crate is being pulled to a distance of 25 m, how much work is done to the crate?

 \(\; \; \; \; \; \quad \quad \text{Given:}\\ \; \; \; \; \; \; \quad \quad \quad m=20 \;kg\\ \; \; \; \; \; \; \quad \quad \quad F=40 \;N\\ \; \; \; \; \; \; \quad \quad \quad d=25\;m\\ \; \; \; \; \; \; \quad \quad \quad \theta=45^\circ\)

We will now use the general equation of work,

\(W=F\cos\theta \cdot d\\ \; \; \; \,=(40\;N)(\cos 45^\circ)(25\;m)\\ \; \; \; \,=(40\;N)(0.71)(25\;m)\\ \; \; \; \,=710 \;J\)

The work done in pulling the crate is 710 J.

4. A 300-kg loaded cart is moved by a 385-N force at an angle of 30° to the ground. How much work is done in moving the cart to 15 m?

\(\; \; \; \; \; \quad \quad \text{Given:}\\ \; \; \; \; \; \; \quad \quad \quad m=300 \;kg\\ \; \; \; \; \; \; \quad \quad \quad F=385 \;N\\ \; \; \; \; \; \; \quad \quad \quad d=15 \;m \\ \; \; \; \; \; \; \quad \quad \quad \theta=30^\circ\)

\(\; \; \; \; \; \; \quad \quad \quad \text{Solution:}\\\)

\(W=F\cos\theta \cdot d\\ \; \; \; \,=(385\;N)(\cos 30^\circ)(15\;m)\\ \; \; \; \,=(385\;N)(0.87)(15\;m)\\ \; \; \; \,=5,024.25 \;J\)