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(I) A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow.What is the skier's change in velocity?
(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?
(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?
(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?
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welcome to our second example video considering restoring and dissipated forces in this video. We'll continue to look at springs, but let's look at them in terms of the math. So let's say we have an object that comes in with some initial velocity V not, and it runs into a spring and then is brought to a stop. We want to find out how much work did the spring do on the object? Well, on the surface, it's pretty easy because we can simply say one half m the final squared minus one half and the initial squared. And if the final is zero, then the work done by the spring is negative one half m be not squared. But what if we want to take this a step further and we say, Well, I also know that work is equal to force D X and now I can set that equal to a Delta K as well. As long as the work The force that is the Spring Force is the Onley force exerted onto the box. Okay, So plugging that in then I have integral of negative k x d x, and I'm going to take that from X equals zero to some negative amount here. So we'll just type in negative X and we can solve for what that might be To find out how far the spring will be compressed by this work. So doing the integral. Then we come up with an negative K times one half X squared, evaluated from zero to negative X. Well, the negative ends up getting squared out. And so we now have work is equal to negative one half k and we could put down Delta X squared, though in this particular case, it will just be X squared because we're starting at X equals zero. So we have negative one half okay, X squared, setting that equal then Teoh what we found up here. Initially we have negative one half K X squared is equal to negative one half m v not squared the one half's cancel. The negatives cancel and X is equal to the square root of em. Over k. I was being not squared. All right, so let's check the units here real quick. We have kilograms divided by kilograms per second squared and that's multiplied by a meter squared per second squared. So when we plug this in. This is an X over here, not a B What we solve for his X kilograms. Cancel seconds. Cancel. We're left with meters squared, we take the square root and we have meters left. So this is in fact, the correct answer. X is going to be equal to the square root of M u R k p not squared. Now you will notice, though, that, um, we did lose the negative here. And so it's not showing us that it was actually and we compressed it a negative distance. Here are a negative displacement. Um, that's okay. There are worse things that could have happened if you're paying attention, remember also that this is equal to plus or minus the square root of, and it is in fact, the minus. In this physical situation, A lot of times when you take a square, do you need to really pay attention in physics class, whether or not you have a plus or a minus, because one will be physical and one will not be physical
Equilibrium and Elasticity
Moment, Impulse, and Collisions
Rotation of Rigid Bodies