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(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?
I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?
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welcome to our section on kinetic energy. Kinetic energy is We'll see here is related to work. Let's go back to our definitions of work. We had to work is equal to magnitude of forced times, magnitude of displacement, cosine of the angle between them and then we also had the integral form. Work is equal to F D X. Now let's consider these two equations and the context of net force Net force is equal to mass times acceleration. So if we plug that in in the integral side, we have integral of mass times acceleration DX. Remember that A is equal to D V D T leaving us with and is a constant So it comes out. We have DVD X over DT Well, DX DT is simply velocity meaning weaken right m times integral of B d. V. If we take the integral of this from the beginning to the end of our condition that well in, we'll end up with one half am the final squared minus one half m the initial squared. So work done by a net force results in a change in velocity or more specifically, in a change of this one half MV Squared quantity which we will end up naming kinetic energy. Okay, so that's kinetic energy. Now we can do the same thing with the algebraic form. If we say that they're pointed in the same direction. When we plug in mass times acceleration, I'm going to use three Kinnah Matic equation V squared equals V not squared, plus to a Delta X note that over here we did not have to make any assumption of a constant force. But over here we are, and we result then in a is equal to V squared minus B not squared over to Delta X When I plugged that. And then I have work is equal to m times Delta X times B squared minus V not squared over to Delta X, Delta X cancels and we have work is equal to one half m. The final squared minus one half m. The initial squared exactly what we found when we did the integral definition. And so both of these things result in the same definition of kinetic energy, which is that kinetic energy equals one half MV squared. More importantly, what it tells us is that work is equal to a change in kinetic energy. Kinetic energy will have the same units is work, so it will also be measured in jewels. And what we'll find is that kinetic energy and this term energy that we use in general, um, will serve as a really nice mathematical model for how things move. And we can consider the energy of objects. Or we can consider the forces acting on the objects. Really, they're just two sides of the same coin where, considering the forces, we'd use things like Newton's second law and dynamics that we have been doing. Whereas if we consider energy, then we'll have a different set of tools at our disposal, as you'll see as we get farther and farther into this unit. So in the next few examples, we're going to consider how to deal with calculating kinetic energy. And just so we have that under our belt, and then we'll start looking at some more advanced topics.
Equilibrium and Elasticity
Moment, Impulse, and Collisions
Rotation of Rigid Bodies