Question Video: Finding the Kinetic Energy of a Body Moving between Two Points given Their Coordinates and the Acting Force in Vector Form | Nagwa Question Video: Finding the Kinetic Energy of a Body Moving between Two Points given Their Coordinates and the Acting Force in Vector Form | Nagwa

Question Video: Finding the Kinetic Energy of a Body Moving between Two Points given Their Coordinates and the Acting Force in Vector Form Mathematics

The coordinates of the points 𝐴 and 𝐵 are (−8, −8) and (9, −3). A body of unit mass moved from 𝐴 to 𝐵 in the direction of 𝐀𝐁 under the action of the force 𝐹, where 𝐅 = (6𝐢 hat + 7𝐣 hat) force units. Given that the body started moving from rest, use the work–energy principle to find its kinetic energy at point 𝐵.

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Video Transcript

The coordinates of the points 𝐴 and 𝐵 are negative eight, negative eight and nine, negative three. A body of unit mass moved from 𝐴 to 𝐵 in the direction of 𝐀𝐁 under the action of the force 𝐹, where 𝐅 is six 𝐢 hat plus seven 𝐣 hat force units. Given that the body started moving from rest, use the work–energy principle to find its kinetic energy at point 𝐵.

The work–energy principle equates the net work done on some object to the total change in kinetic energy of that object. The total change in kinetic energy of an object is the difference between its final kinetic energy and its initial kinetic energy. In this case though, our object starts from rest, so its initial kinetic energy is zero. But if the initial kinetic energy is zero, then the total change in kinetic energy is just the final kinetic energy, which is, in fact, the kinetic energy that we’re looking for. Okay, well, that means that to find the final kinetic energy that we’re looking for, all we need to do is evaluate the left-hand side of the work–energy principle, that is, find the net work done to the object.

Work itself is the energy gained or lost by an object as it moves in the presence of some external force. The simplest expression for work is force times distance, which is valid as long as the force is parallel to the object’s motion. In our situation though, we’re given a vector representing the force. And we’re also told the beginning and endpoints of the object’s motion and also that the object moved in a straight line between these two points. So if we want to use force times distance, we need to make sure that the force vector and the direction of motion are indeed parallel. Here, we have a set of axes and we’ve labeled the initial and final points 𝐴 and 𝐵, respectively.

The vector 𝐀𝐁, with its tail at point 𝐴 and head at point 𝐵, is the vector that the object follows as it moves. We call this the displacement vector. The displacement vector for any motion has its tail at the initial position and its head at the final position and represents the net motion of the object. Also, on this diagram is the force vector 𝐅 six 𝐢 hat plus seven 𝐣 hat. Just by looking at this diagram, which we have carefully drawn to scale, we can see that 𝐅 and 𝐀𝐁 are not parallel. This means that we’ll need a different way to express the work. The formula we’ll use involves the vector dot product of the force and the displacement, specifically work equals force dotted with displacement.

This formula is valid as long as the force is constant for the entire motion. And in fact, six 𝐢 hat plus seven 𝐣 hat is constant. Since force dot displacement gives us the work and work gives us the kinetic energy that we’re looking for, all we need to do is calculate the dot product of 𝐅 and 𝐀𝐁, the displacement of the motion. Of course, to carry out this dot product, we first need to find an expression for the vector 𝐀𝐁. To find the vertical and horizontal components of this vector, we’ll need to consider the changes in the vertical and horizontal coordinates between the points 𝐴 and 𝐵. The vertical change is a rise from negative eight to negative three. This is a total change of positive five, and it is in the 𝐣 hat direction.

The total horizontal change is the change from negative eight to positive nine. This is a net change of 17, and it is in the 𝐢 hat direction. So the vector 𝐀𝐁 itself expressed in the unit vectors 𝐢 hat and 𝐣 hat would be 17𝐢 hat plus five 𝐣 hat.

All right, now let’s calculate the dot product. To calculate the dot product of two vectors, we multiply the components with the same unit vector and then add those results together. So in this case, the dot product would be six times 17, the product of the 𝐢 hat components, plus seven times five, the product of the 𝐣 hat components. Six times 17 is 102, and seven times five is 35. 102 plus 35 is 137. And this is the numerical value of the total work done on the object. Now we need to give this answer units. The units that we have for force are the rather arbitrary force units, which just means whatever unit we use for force. Therefore, our answer will be 137 work units because it’s the calculation of work that we did from force measured in the arbitrary force units.

What this really means is just that the unit of force determines the unit of work. For example, if we decide that the force was measured in newtons, then the work unit would be joules. But if we used any other unit for force, we would still have the same numerical answer, just a different unit for the work. Now, remember that the work energy theorem tells us that the final kinetic energy that is what we’re looking for is exactly the net work done on the system. So the final kinetic energy is 137 work units. It’s also worth mentioning that the work energy theorem guarantees that work units are exactly the same as energy units.

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