Work and Energy

In physics, the terms work and energy have very specific meanings that are different from what we refer to in our daily lives.

In other words, work done is the action on an object, which, in turn, moves an object in a certain direction. Pushing a box across a floor is an example of work done.

Figure 1. A force applied that moves an object in the direction of the force applied creates work.

How can we calculate the work done and what are its units?

We can calculate work done on an object using the following formula:

\[W = F \cdot d\]

Here:

• W = work done.
• F = force applied.
• d = distance covered in the direction of the force applied.

Work has the SI unit of Joules (Nm). According to the formula, work can only be done if the object on which the force is applied moves in the same direction as the force. If you apply force on an object and there is no displacement, you have done no work regardless of how tired you feel.

What if the force applied does not move an object in the direction of the force?

While calculating the work done on an object, we must also bear in mind that only the component of the force that is parallel to the direction of the movement is taken into consideration. If you apply a force on a box, and your arms are parallel to the floor, all your efforts will contribute to the work done on the box.

Figure 2. Only the component of the force that is parallel to the movement of the object is the force that results in work. Source: Oğulcan Tezcan, Vaia.

If, on the other hand, your arms are not parallel to the floor and you are pushing the box at an angle, only the component of that force that is parallel to the movement will be taken into account while calculating the work done. The component of the applied force that is perpendicular to the direction of the motion is discarded.

In situations where the force applied is at an angle to the displacement and we have to calculate the horizontal distance covered, we use the following formula:

\[W = F \cdot d \cos(\theta)\]

Here, θ is the angle from the horizontal plane.

In figure 2, if the angle theta is zero, the applied force is completely parallel to the direction of motion, and the equation becomes W = F⋅d. If theta is equal to 90° and the direction of the force is downwards, the equation results in zero and the box will not move, as the applied force will be pushing the box into the ground.

If the vertical force happens to be in the upward direction, the box will move in the upward direction (if the force is great enough), and work will be done.

Can there be both positive and negative work?

You might be thinking that work has a force and makes an object move in a certain direction, but this is not the case. Work is a dot product of two vector quantities: force and displacement. And as the dot product of two vector quantities is a scalar quantity, work also becomes a scalar quantity. It has a magnitude but no specific direction.

Just like temperature is a scalar quantity but can have negative values, work can also be positive or negative.

Work is positive if the force applied to an object is in the same direction as the displacement. Work is negative if the applied force is in the opposite direction to the movement of the object.

Energy and its units

Previously, we defined the work done as the energy transferred to an object, which indicates that the energy element plays an important role in determining the total work done.

Energy, which determines how much work a body can do, is measured in Joules (Nm). Energy is merely a property of a system that can be transferred to another system.

Concept of Work and Energy

This theorem states that whenever work is done on an object, there will be a change in the kinetic energy of that object. Kinetic energy happens to be the energy of motion. But how can we prove this theorem?

First of all, we are going to make a few assumptions, even though these assumptions are not required to derive the formula to prove the work-energy theorem. They just make the calculations much simpler.

Suppose you have forces acting on an object whose direction is the same as the displacement. Theta will be zero in this case, which reduces the equation to:

\[W = F \cdot d\]

Figure 3. Net force in the direction of the movement.

As we are dealing with the W_net total work on an object, we can rewrite the above formula as:

\[W_{net} = F_{net} \cdot d\]

F_net is the total net force applied to an object. We also know that the net force is equal to the product of mass and acceleration. Hence:

\[F_{net} = ma\]

\[W_{net} = m \cdot a \cdot d\]

The above equation has acceleration, which is why we need to rewrite it in the velocity v form. Recall from the kinematics equations that:

\[v^2_f = v^2_i + 2 \cdot a \cdot d\]

\[a \cdot d = \frac{v^2_f - v^2_i}{2}\]

V_f in the above equation is the final velocity, V_i is the initial velocity, a is the acceleration, and d is the displacement. To use the above kinematic equation, we need to assume the acceleration to be constant. In that case, the F_net applied is constant as well.

Putting the values into the W_net equation yields:

\[W_{net} = m \cdot a \cdot d\]

\[a \cdot d = \frac{v^2_f - v^2_i}{2}\]

\[W_{net} = m\Big(\frac{v^2_f - v_i^2}{2} \Big) = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2\]

This quantity of \(\frac{1}{2} \cdot m \cdot v^2\) is what we call kinetic energy. Finally, we can write the work-energy theorem as:

\[W_{net} = \Delta KE\]