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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?
Suman Saurav T.
(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?
(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?
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welcome to our second example for one dimensional dynamics in this video will consider the physics of a sled. Try that again, being hauled by a dog. Okay, alright, so sled being hauled by a dog things everything's horizontal. So forces air being applied in along the same axis will say that the sled has a mass equal to 40 kg and the dog has a mass equal to 20 kg. And we want to see what's it going to take for the dog to pull this? Let along. Now, Um, when we draw the free brought diagrams for this, remember, we only want to consider one direction. Obviously, there are forces in the why, but we're not going to worry about those so much a massive dog masters that. So we're going to say that the force due to the dog call it a force of tension because you see that it's tied to the dog. Meanwhile, the dog will experience a force of tension backwards in the opposite direction. His Now, if we assume that the there's no friction force on the sled, then there won't be any other forces here. Meanwhile, for the dog, we need a force of friction, pushing it for pushing it forward. So we have The dog is experiencing some force of friction. We'll just call it a force. So Beth forward. And when we write our equations of motion, we have F T is equal to massive sled times its acceleration. Over here we have force friction minus F T is equal to mass of dog times its acceleration. Now, as we've seen before in this example, what we can do then is we can plug in. We know the force of tension so we can just plug that straight in its equal to massive sled times. Acceleration is equal to massive dog times acceleration, and we find that the force pushing the dog forward has to be equal to the mass of the sled, plus the mass of the dog all times acceleration. So it looks like the force that's pushing the dog for it has to be acting on both the sled and the dog as if they were a single object. This is the same result we found for pushing eso. That shouldn't come as any surprise. It's just something we need to make sure we keep track of here again. Once we get into friction, we can consider the nuances there. This problem will become a little more complex and interesting. But for now, this is all that it requires in our frictionless, more or less except for the force pushing the dog forward. I am this frictionless environment.
Equilibrium and Elasticity