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Ch 04: Newton's Laws of Motion

Chapter 4, Problem 4

A 4.50-kg experimental cart undergoes an acceleration in a straight line (the x-axis). The graph in Fig. E4.13 shows this acceleration as a function of time. (b) During what times is the net force on the cart a constant?

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Hey everybody. So today we're being told that we have an 8.5 kg trolley that is accelerating in a straight line in the X axis direction. Now the acceleration has been plotted as a function of time in the graph below. And we're being asked to determine the times when the trolley experiences a constant net force. Now, before determining this, I'd like to take a look and do some dimensional analysis to sort of analyze the area under the graph all over here and I want to take special note of this because the area under a graph can usually tell us some very important things sometimes. So if we are to analyze, let's just take a small portion right here. If we take this rectangle well to find the area, we'd multiply the X axis by the y axis because that is based on tight with the X axis by the Y axis. So that means we would multiply acceleration which is in meters per second squared and our time, which let's just say is in seconds for now, which means we would get the area as meters per second, which is the same thing as velocity. So this means that the area under a acceleration versus time graph will be equal to the velocity. However, for the sake of this problem, we don't need velocity here because we're trying to find the Uh force here and recall that according to Newton's 1st Law, net force is equal to mass into acceleration, which means we have no use for velocity here. But we're just covering our bases with this in mind though. If we want a constant, if we want a constant force, if constant force must be achieved, well, that means that both the mass and the acceleration must stay constant. Well, the mass will never change. We're not talking about changing or adding anything onto the trolley. It already has a massive 8.5 kg. So we can rule that out, which means we need to find a place or a time period on the graph in which the acceleration is constant and we can see that between times one and seven seconds. Because the acceleration here Stays at to stays at 2m/s square, which means it is constant. During that time period, it doesn't change. So this means the times when f is constant is when the acceleration is constant or between Times 1/2 and seven seconds. And that corresponds with answer choice F. I hope this helps. And I look forward to seeing you all in the next one.
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