Video Transcript
An object with a mass of 22 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? Answer to the nearest newton.
This question presents us with a graph that shows the change in an object’s velocity while a force is applied to it. We can recall that the rate of change of an object’s velocity gives us the acceleration of the object and that this is equal to the slope or gradient of a line on a velocity–time graph, like the one that we’re given in this question. So, by finding the slope of this line, we can calculate the acceleration of the object. We might wonder why we would want to find the object’s acceleration when the question is actually asking us to work out the force applied to the object. The answer to this comes from Newton’s second law of motion.
This law says that the force acting on an object is equal to the mass of the object multiplied by the object’s acceleration. The law is often expressed using symbols as force 𝐹 is equal to mass 𝑚 multiplied by acceleration 𝑎. The question tells us the object’s mass; it’s 22 kilograms. This means that we know the value of 𝑚. So, if we can work out the value of the acceleration 𝑎, we can use 𝑚 and 𝑎 to calculate the force 𝐹 acting on the object. We’ve said that to calculate the acceleration, we need to find the slope of the velocity–time graph.
The slope of a graph is the change in the vertical coordinate over some section of the graph divided by the change in the horizontal coordinate for the same section. In this case, since it’s a velocity–time graph, the vertical coordinate is velocity and the horizontal coordinate is time. So, the slope of the graph is equal to the change in velocity Δ𝑣 divided by the change in time Δ𝑡. And this will give us the acceleration 𝑎 of the object. Let’s consider the whole of the graph, that is, between this point at the far left and this point at the far right.
We can see that the far-left point is at a time of zero seconds. Meanwhile, we find that the point on the right is at a time of five seconds. So, the time interval between these two points, Δ𝑡, is equal to five seconds, the time at the right-hand point, minus zero seconds, the time at the left-hand point. This gives us that Δ𝑡 is equal to five seconds. Now, looking at the velocity values, we can see that the left- hand point has a velocity of five meters per second. Meanwhile, tracing across from the right-hand point, we find that this point has a velocity of 20 meters per second. So, the change in velocity between the points Δ𝑣 is equal to 20 meters per second, that’s the velocity of the right-hand point, minus five meters per second, the velocity of the left-hand point. When we evaluate this, we find that Δ𝑣 is equal to 15 meters per second.
So, the slope of this velocity–time graph, which gives us the acceleration 𝑎 of the object, is equal to 15 meters per second, that’s the value of Δ𝑣, divided by five seconds; that’s the value of Δ𝑡. When we evaluate this expression, we find that the acceleration 𝑎 is equal to three meters per second squared.
So, we have found the object’s acceleration. And we know from the question that the object has a mass 𝑚, which is equal to 22 kilograms. We can now take these values for 𝑚 and 𝑎 and sub them into this equation from Newton’s second law of motion to calculate the force 𝐹 applied to the object. When we do this, we get that 𝐹 is equal to 22 kilograms, that’s the value of 𝑚, multiplied by three meters per second squared; that’s the value of 𝑎. We can notice that the mass is in units of kilograms, the SI base unit for mass, and the acceleration is in units of meters per second squared, the SI base unit for acceleration. This means that the force 𝐹 that we will calculate will be in the SI base unit for force, which is the newton.
Evaluating this expression gives us a result of 66 newtons. And this value is the force that’s applied to the object, which is what the question was asking us to find. We can notice that the question asks 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 of how much force is applied to the object is 66 newtons.
