### Video Transcript

An object of mass 75 kilograms has a force applied to it. The graph shows the change in the object’s velocity while the force is applied. How much force is applied to the object? Give your answer to the nearest newton.

This question gives us a graph, and that graph shows the change in an object’s velocity while a force is applied to it. The graph plots velocity on the vertical axis against time on the horizontal axis. And so, the slope of the graph, which is defined as the change in the vertical coordinate divided by the change in horizontal coordinate, is equal to the change in velocity, Δ𝑣, divided by the change in time, Δ𝑡. We can recall that the acceleration of an object is equal to the rate of change of that object’s velocity. And the rate of change of velocity will be equal to the change in velocity divided by the change in time over which that velocity change occurs. This means that the acceleration 𝑎 of an object is equal to the slope of that object’s velocity–time graph.

Now, the question isn’t actually asking us to find the acceleration, but rather the force that is applied to the object. However, we’re going to see that it will be useful to first find this acceleration in order to then work out the force. The reason for this is Newton’s second law of motion. This says that the force applied to an object is equal to the object’s mass multiplied by its acceleration. This is often written in terms of symbols as force 𝐹 is equal to mass 𝑚 multiplied by acceleration 𝑎. The question tells us that the mass of the object is 75 kilograms. So, in this equation, we know the value of 𝑚. This means that if we can work out the value of the acceleration 𝑎, then we have all of the information that we need to calculate the force applied to the object.

So, let’s work out the value of 𝑎 by finding the slope of this velocity–time graph. We’ll consider the two points that are right at the ends of the graph. So, that’s this point here on the left and this point on the right. We can see that the left-hand point is at a time value of zero seconds. Meanwhile, tracing down from the right-hand point until we get to the time axis, we see that this point has a time value of five seconds. The change in the time value between this point and this point on the graph, which is Δ𝑡, is equal to five seconds, so that’s the time value at the right-hand point, minus zero seconds, the time value at the left-hand point. This gives us that Δ𝑡 is equal to five seconds.

Next, we’ll look at the velocity values. We can see that the left-hand point has a velocity value of zero meters per second. Then, looking now at the right-hand point, we see that it traces across to this height on the velocity axis. This height is one-fifth of the way between the 12 meters per second mark and the next mark, which would be 14 meters per second. This gives it a value of 12.4 meters per second. Then, the change in velocity Δ𝑣 between the two points on the graph is equal to 12.4 meters per second, that’s the velocity at the right-hand point, minus zero meters per second, the velocity at the left-hand point. This gives us that Δ𝑣 is equal to 12.4 meters per second.

We can now sub these values for Δ𝑣 and Δ𝑡 into this expression. This lets us calculate the slope of the graph, which gives us the acceleration 𝑎 of the object. Subbing in our values gives us that 𝑎 is equal to 12.4 meters per second divided by five seconds. Then, evaluating the expression, we find that 𝑎 is equal to 2.48 meters per second squared. So, we know the mass 𝑚 of the object, and we have now found the object’s acceleration 𝑎. We can now take these values and sub them into this equation to calculate the force 𝐹 that’s applied to the object. Subbing the values, we get that 𝐹 is equal to 75 kilograms, that’s the value for 𝑚, multiplied by 2.48 meters per second squared, the value for 𝑎.

At this point, it’s worth noticing that the mass is expressed in units of kilograms, the SI base unit for mass, and the acceleration is in units of meters per second squared; that’s the SI base unit for acceleration. Since both of these two quantities are expressed in their SI base units, this means that the force we’re going to calculate will be in the SI base unit for force, which is the newton. Evaluating this expression, we calculate a force of 186 newtons. This value is the force applied to the object, which is what we were asked to find.

We can notice that the question asked us to give our answer to the nearest newton. But our result is already an integer number of newtons, so we don’t need to do anything further. This means that our final answer to the question is that the amount of force applied to the object is 186 newtons.