In this explainer, we will learn how to calculate the work done by a constant force vector acting on a body over a displacement vector using the dot product.
The work done on a body by a force can be defined as follows.
Definition: Work Done on a Body by a Force
The work done on a body by a force is dependent on the force that acts on the body and the distance that the body moves in the direction of that force according to the formula where is the magnitude of the force, is the magnitude of displacement of the body while the force acts on it, and is the angle between the directions of and .
An alternative way of representing the work done on a body by a force is to represent the force and the displacement as vectors rather than as the magnitudes of vectors.
The product of two vectors and can be the dot product of the vectors, which is defined as follows.
Definition: Dot Product of Two Vectors
The dot product of two vectors is given by where is the angle between and . The angle is taken counterclockwise from to , as shown by the following figure.
The work done by a force with magnitude over a displacement with magnitude is equal to
A graphical representation of and demonstrates that the product of the magnitude of and the magnitude of the component of in the direction of is equal to .
The dot product of two vectors expressed in component form can be determined without referring to the angle between the vectors.
Suppose now that vectors and are perpendicular, as shown in the following figure.
The product is given by
The scalar product of two perpendicular vectors is zero. There is no force acting on the body in the direction of the displacement, and so the force does no work on the body.
Let us look at an example of using vector notation to determine the work done by a force.
Example 1: Calculating the Work Done by a Force Acting on a Particle Where the Force and Position Are Given as Vectors
A particle moves in a plane in which and are perpendicular unit vectors. A force acts on the particle. The particle moves from the origin to the point with position vector m. Find the work done by the force.
The work done by the force is the dot product of the force vector and the displacement vector of the particle. The question does not give a displacement vector, however, it gives a position vector. The question also states that the particle moves to the stated position from the origin, and so the displacement vector of the particle is given by which is equal to the given position vector.
The work done by the force, , is given by the dot product of the vectors, which is given by
The work done is therefore given by
The work done is negative. If the energy of the particle is conserved, then the kinetic energy of the particle must decrease. If the energy of the particle is not conserved, the work done might instead increase the potential energy of the particle.
Now let us consider an example where multiple forces act on a body to produce a displacement.
Example 2: Finding the Work Done by the Resultant of Two Forces Given Acting on a Body
A body moves in a plane in which and are perpendicular unit vectors. Two forces and act on the body. The particle moves from the point with position vector m to the point m. Find the work done by the resultant of the forces.
Two forces act on the body. The forces are vectors and the resultant of the vectors can be determined by summing the components of the vectors. The -component of the resultant force is given by and the -component of the resultant force is given by
The resultant force on the body is therefore
The final position vector of the body is and the initial position vector of the body is
The displacement vector from the initial position to the final position is therefore
Using the equation for finding the work done, we have that which gives us
Let us now look at an example where neither the components of the force vector nor the components of the displacement vector are directly given.
Example 3: Using Vectors to Find the Work Done by a Force with the Direction and Magnitude Given Separately
A particle moved from point to point along a straight line under the action of a force of magnitude N acting in the same direction as the vector . Calculate the work done by the force, given that the magnitude of the displacement is measured in metres.
We can call the displacement vector, , . Vector is shown with vector in the following figure.
Vector is given by
The dot product of and does not equal the dot product of and .
The magnitude of is given, but to use in a dot product calculation, the and components of must be determined.
Because acts along the line of , the component of must have a magnitude that is equal to three times the magnitude of the component of . This relationship can be expressed as
and are two sides of a right triangle with a hypotenuse that has a length of . It is therefore the case that
And again, as
Knowing the components of , we can express as
The work done, , is therefore
Now let us look at another example where neither the components of the force vector nor the components of the displacement vector are directly given.
Example 4: Finding the Work Done by a Force in Vector Form Acting on a Body Moving between Two Points
A particle moved from point to point along a straight line under the action of the force acting in the opposite direction to the displacement . Find the work done by the force .
The displacement vector is the vector that has its tail at and its head at . The vector is given by
The direction of is . The direction in which acts is opposite to , so must act along a line
The fraction of the -component over the -component is given by
In order for to have the same direction as , the fraction of the -component over its -component must be equal to that of . Therefore, we have that
Rearranging to make the subject of the equation gives
We can now express as
The work done, , is therefore
Unsurprisingly, the work done is negative, as the force acts in the opposite direction to the displacement.
When the energy of a particle that is acted on by a force is conserved, the path that the particle travels to displace it while the force acts on it does not affect the work done on the particle by the force.
Now let us look at an example where the work done by a force in a time interval is considered.
Example 5: Finding the Work Done by a Force when the Displacement is Given in Terms of Time
A particle moves in a plane in which and are perpendicular unit vectors. Its displacement from the origin at time seconds is given by and it is acted on by a force . How much work does the force do between and ?
The inclusion in this example does not require it to be solved in a way that greatly differs from the previous examples, as dot products of vectors are still used in the same way. The only novel feature found in this example is that we should find the displacement that correspond to two different times.
The displacement of the particle at and must be determined. This is done by substituting 3 and 2 as the values of to give and , respectively, which gives us
The displacement of the particle between and is minus , given by
The dot product of and gives us the work done, , between and . Since this is given by
- The work done by a constant force over a displacement is equal to the dot product of and , or where is the angle between and .
- The dot product of and can be determined without use of the angle by expressing and in component form.
- If the components of and are in the perpendicular directions and , then