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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$^2$ ?
(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?
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Welcome to our final example video Looking at drag forces in this video, we're going to return to the solution we found when considering after AG is equal to a constant times velocity. Remember, we found this only works for small objects moving very slowly. But say it. It is true. Here we found that we had a velocity is a function of time was equal to terminal velocity multiplied by one minus e to the negative. Que over em times t Okay. Remember, Terminal velocity in this case is equal to M G over K. And so we can use this to plug in there. Alright, Eso questions we might ask What are the acceleration in the position? So to find acceleration, we say the derivative of velocity is function of time. With respect to time is going to be equal to well, the constant goes away and we're left with derivative respect Time of negative terminal velocity E to the negative k over em times t which, uh, it is an exponents, so that's pretty easy. The negatives will end up canceling and will have VT times k over em times e to the negative. Que over em Times t notice when t equals zero That that means the acceleration then would be equal to V T. K over him. So this is the initial acceleration of the object. Um, for this particular case, remember, we're considering the case of we have m g down and drag equal to cave. All right, so there's our acceleration. Now, if we wanted to find, uh, I should write down officially, then a f t is equal to V T. Okay, over m, which is just our initial acceleration times e to the negative. Okay, over m times t. So acceleration is a plot would look something like this would get smaller and smaller and smaller actually would never cross the x axis. It would just approach it. Okay, um, if we want to find position as a function of time, then you would say x of t is equal to the integral of V f t DT. And you would find I guess we should say why of tea since we're going up and down. So why not that it it really is that important? We could have taken the X axis and pointed it up. But why of tea then is going to be equal to terminal velocity Times T minus mm. Over K. Times one minus one minus E to the negative k over em. Times T. Okay, so we have two separate sets of parentheses here on inner one for one minus E to the negative K. Um, over T. And then an outer one that puts it in with this term, this extra term t um Okay. And so now we have our position as a function of time and we could go through and Saul for where it will be at any particular time so you can do the same thing with the other drag equation. It's a little more challenging, but if you've taken a differential equations course, you should be able to find the solution without any issue.
Equilibrium and Elasticity