An object moves along a straight line. As the object moves, it is acted upon by a constant force. The diagram shows the displacement of the object, 𝐝, and the force acting on it, 𝐅. Each of the grid squares in the diagram has a side length of one in both meters and newtons. Calculate 𝐅 dot 𝐝.
This question presents us with a diagram showing two vectors labeled 𝐝 and 𝐅. The question tells us that 𝐝 is the displacement of an object and 𝐅 is the force acting on that object. We’re also told that this force 𝐅 is constant, which means that it’s not changing with time. The question asks us to work out the scalar product 𝐅 dot 𝐝. So, this is the scalar product of the force 𝐅 acting on our object and the displacement 𝐝 of that object.
We can recall that the scalar product of force and displacement gives us the work done on the object by the force. If the force is measured in units of newtons and the displacement is measured in meters, then the scalar product of force and displacement will have units of newton meters, which is equivalent to joules, the unit of energy. Since this question is asking us to calculate a scalar product, we should also recall the definition of the scalar product of two vectors.
We’ll consider two general vectors that we’ll label 𝐀 and 𝐁. If we suppose that both of these vectors lie in the 𝑥𝑦-plane, then we can write them in component form as an 𝑥-component labeled with a subscript 𝑥 multiplied by 𝐢 hat plus a 𝑦-component labeled with a subscript 𝑦 multiplied by 𝐣 hat. Remember that 𝐢 hat is the unit vector in the 𝑥-direction and 𝐣 hat is the unit vector in the 𝑦-direction. Then the scalar product 𝐀 dot 𝐁 is given by the 𝑥-component of 𝐀 multiplied by the 𝑥-component of 𝐁 plus the 𝑦-component of 𝐀 multiplied by the 𝑦-component of 𝐁. In other words, the scalar product of two vectors is given by the product of the vectors’ 𝑥-components plus the product of their 𝑦-components.
This general expression for the scalar product of two vectors tells us that if we want to work out the scalar product 𝐅 dot 𝐝, then we need to find the 𝑥- and 𝑦-components of our vectors 𝐅 and 𝐝. These vectors are given to us in the form of arrows drawn on a diagram. And the question tells us that the squares in this diagram have a side length of one in both meters and newtons. This means that when we’re talking about the vector 𝐝, then one square in the diagram means one meter, and when we’re talking about 𝐅, then one square is one newton.
If we add a set of axes to our diagram with the origin positioned at the tail of the two vectors, then we can easily read off the number of squares that each vector extends in the 𝑥-direction and the 𝑦-direction. Since we know that for our vector 𝐝 each square corresponds to one meter and that for 𝐅 each square corresponds to one newton, then the number of squares that each vector extends in the 𝑥- and 𝑦-directions directly gives us the 𝑥- and 𝑦-components of that vector.
So, let’s count the number of squares that each of our vectors extends. We’ll begin with vector 𝐅. We see that 𝐅 extends two squares in the positive 𝑥-direction and four squares in the positive 𝑦-direction. Since for the vector 𝐅, one square corresponds to one newton, then we know that the 𝑥-component of 𝐅 is two newtons and the 𝑦-component is four newtons. So, in component form, 𝐅 becomes two newtons 𝐢 hat plus four newtons 𝐣 hat.
Now, we’ll do the same thing for the vector 𝐝. We see that 𝐝 extends eight squares in the positive 𝑥-direction and three squares in the positive 𝑦-direction. For the vector 𝐝, one square corresponds to one meter. So, the 𝑥-component of 𝐝 is eight meters and the 𝑦-component is three meters. In component form, we have that 𝐝 is equal to eight meters 𝐢 hat plus three meters 𝐣 hat.
Now that we have both of our vectors 𝐅 and 𝐝 written in component form, we’re ready to calculate the scalar product 𝐅 dot 𝐝. Looking at our general expression for the scalar product, we see that the first term is given by the product of the 𝑥-components of the vectors. So, for our scalar product 𝐅 dot 𝐝, that’s the 𝑥-component of 𝐅, which is two newtons, multiplied by the 𝑥-component of 𝐝, which is eight meters. Then, we add to this a second term given by the product of the 𝑦-components. In our case, that’s the 𝑦-component of 𝐅, which is four newtons, multiplied by the 𝑦-component of 𝐝, which is three meters.
All that’s left to do now is to evaluate this expression here. The first term is two newtons multiplied by eight meters, which gives 16 newton meters. And the second term is four newtons multiplied by three meters, which gives 12 newton meters. Adding together 16 newton meters and 12 newton meters gives the result of 28 newton meters.
Now, we said already that when we calculate the scalar product 𝐅 dot 𝐝, we get the work done on the object by the force. This work done has units of energy, and specifically when the force is measured in newtons and the displacement is measured in meters, then the work done has units of joules since one newton meter is equal to one joule. This means that we can take our result of 28 newton meters and write this as 28 joules. And so, our final answer is that the scalar product 𝐅 dot 𝐝, which gives us the amount of work done by the force 𝐅 on the object, is equal to 28 joules.