Lesson Explainer: Work Done by a Force Expressed in Vector Notation Mathematics

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 𝑊=𝐹𝑑(𝜃),cos where 𝐹 is the magnitude of the force, 𝑑 is the magnitude of displacement of the body while the force acts on it, and 𝜃 is the angle between the directions of 𝐹 and 𝑑.

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 𝑎 and 𝑏 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 𝑎𝑏=𝑎𝑏(𝜃),cos where 𝜃 is the angle between 𝑎 and 𝑏. The angle is taken counterclockwise from 𝑎 to 𝑏, as shown by the following figure.

The work done by a force with magnitude 𝐹 over a displacement with magnitude 𝑑 is equal to 𝐹𝑑𝐹𝑑=𝐹𝑑(𝜃).cos

A graphical representation of 𝐹 and 𝑑 demonstrates that the product of the magnitude of 𝐹 and the magnitude of the component of 𝑑 in the direction of 𝐹 is equal to 𝐹𝑑(𝜃)cos.

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 𝐹 and 𝑑 are perpendicular, as shown in the following figure.

The product is given by 𝐹𝑑=(4,3)(3,4)𝐹𝑑=(4×3)+(3×4)=1212=0.

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 𝑖 and 𝑗 are perpendicular unit vectors. A force 𝐹=9𝑖+𝑗N acts on the particle. The particle moves from the origin to the point with position vector 9𝑖+6𝑗 m. Find the work done by the force.

Answer

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 𝑑=(90)𝑖+(60)𝑗,m which is equal to the given position vector.

The work done by the force, 𝑊, is given by the dot product of the vectors, which is given by 𝑊=(9,1)(9,6)𝑊=(9×9)+(1×6)=81+6=75.

The work done is therefore given by 𝑊=75.J

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 𝑖 and 𝑗 are perpendicular unit vectors. Two forces 𝐹=9𝑖2𝑗N and 𝐹=9𝑖7𝑗N act on the body. The particle moves from the point with position vector 6𝑖+2𝑗 m to the point 2𝑖+3𝑗 m. Find the work done by the resultant of the forces.

Answer

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 𝑥-component of the resultant force is given by 𝐹=9𝑖+9𝑖=18𝑖,R and the 𝑦-component of the resultant force is given by 𝐹=2𝑗7𝑗=9𝑗.R

The resultant force on the body is therefore 𝐹=18𝑖9𝑗.RN

The final position vector of the body is 2𝑖+3𝑗, and the initial position vector of the body is 6𝑖+2𝑗.

The displacement vector from the initial position to the final position is therefore 𝑑=2𝑖+3𝑗6𝑖+2𝑗𝑑=2(6)𝑖+(32)𝑗=8𝑖+𝑗.mm

Using the equation for finding the work done, we have that 𝑊=𝐹𝑑,R which gives us 𝑊=(18,9)(8,1)𝑊=(18×8)+(9×1)=1449=135.J

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 𝐴(7,3) to point 𝐵(9,2) along a straight line under the action of a force 𝐹 of magnitude 810 N acting in the same direction as the vector 𝑐=3𝑖𝑗. Calculate the work done by the force, given that the magnitude of the displacement is measured in metres.

Answer

We can call the displacement vector, 𝐴𝐵, 𝑑. Vector 𝑑 is shown with vector 𝑐 in the following figure.

Vector 𝑑 is given by 𝑑=(97)𝑖+(2(3))𝑗=16𝑖+5𝑗.m

The dot product of 𝑑 and 𝑐 does not equal the dot product of 𝑑 and 𝐹.

The magnitude of 𝐹 is given, but to use 𝐹 in a dot product calculation, the 𝑖 and 𝑗 components of 𝐹 must be determined.

Because 𝐹 acts along the line of 𝑐, the 𝑖 component of 𝐹 must have a magnitude that is equal to three times the magnitude of the 𝑗 component of 𝐹. This relationship can be expressed as 𝐹=3𝐹.

𝐹 and 𝐹 are two sides of a right triangle with a hypotenuse that has a length of 810. It is therefore the case that 𝐹+𝐹=810=640.

As 𝐹=3𝐹,3𝐹+𝐹=6409𝐹+𝐹=640𝐹=64𝐹=8.

And again, as 𝐹=3𝐹,𝐹=24.

Knowing the components of 𝐹, we can express 𝐹 as 𝐹=24𝑖8𝑗.N

The work done, 𝑊, is therefore 𝑊=(24,8)(16,5)𝑊=(24×16)+(8×5)=38440=344.J

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 𝐴(2,2) to point 𝐵(6,10) along a straight line under the action of the force 𝐹=𝑘𝑖6𝑗 acting in the opposite direction to the displacement 𝐴𝐵. Find the work done by the force 𝐹.

Answer

The displacement vector 𝑑 is the vector that has its tail at 𝐴 and its head at 𝐵. The vector 𝑑 is given by 𝑑=(6(2))𝑖+(10(2))𝑗=8𝑖+12𝑗.m

The direction of 𝑑 is 𝐴𝐵. The direction in which 𝐹 acts is opposite to 𝐴𝐵, so 𝐹 must act along a line 𝐵𝐴𝐵𝐴=8𝑖12𝑗

The fraction of the 𝑦-component over the 𝑥-component is given by Δ𝑦Δ𝑥=128=32.

In order for 𝐹 to have the same direction as 𝐵𝐴, the fraction of the 𝑦-component over its 𝑥-component must be equal to that of 𝐵𝐴. Therefore, we have that Δ𝑦Δ𝑥=6𝑘=32.

Rearranging to make 𝑘 the subject of the equation gives 3𝑘=6×2𝑘=4.

We can now express 𝐹 as 𝐹=4𝑖6𝑗.N

The work done, 𝑊, is therefore 𝑊=(4,6)(8,12)𝑊=(4×8)+(6×12)=3272=104.J

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 𝑖 and 𝑗 are perpendicular unit vectors. Its displacement from the origin at time 𝑡 seconds is given by 𝑟=2𝑡+7𝑖+(𝑡+7)𝑗m and it is acted on by a force 𝐹=6𝑖+3𝑗N. How much work does the force do between 𝑡=2s and 𝑡=3s?

Answer

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 𝑡=3s and 𝑡=2s must be determined. This is done by substituting 3 and 2 as the values of 𝑡 to give 𝑑 and 𝑑, respectively, which gives us 𝑑=2×3+7𝑖+(3+7)𝑗=25𝑖+10𝑗,𝑑=2×2+7𝑖+(2+7)𝑗=15𝑖+9𝑗.mm

The displacement of the particle between 𝑡=3s and 𝑡=2s is 𝑑 minus 𝑑, given by 𝑑𝑑=Δ𝑑=25𝑖+10𝑗15𝑖+9𝑗=10𝑖+𝑗.m

The dot product of 𝐹 and Δ𝑑 gives us the work done, 𝑊, between 𝑡=3s and 𝑡=2s. Since 𝐹=6𝑖+3𝑗,N this is given by 𝑊=(6,3)(10,1)𝑊=(6×10)+(1×3)=63.J

Key Points

  • The work done by a constant force 𝐹 over a displacement 𝑑 is equal to the dot product of 𝐹 and 𝑑, 𝑊=𝐹𝑑, or 𝑊=𝐹𝑑(𝜃),cos where 𝜃 is the angle between 𝐹 and 𝑑.
  • The dot product of 𝐹 and 𝑑 can be determined without use of the angle 𝜃 by expressing 𝐹 and 𝑑 in component form.
  • If the components of 𝐹 and 𝑑 are in the perpendicular directions 𝑖 and 𝑗, then 𝐹𝑑=𝐹,𝐹𝑑,𝑑=(𝐹𝑑)+𝐹𝑑.

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