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07:43

Kathleen T.

(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?

03:09

Aditya P.

(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?

01:40

Keshav S.

(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?

04:27

Kai C.

(I) A 110-kg tackler moving at 2.5 ms meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

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welcome to our third example Video considering restoring and dissipated forces in this video will start to look at friction. Consider a box that is moving at some velocity V not. And then some time later has come to the final equal to 0 m per second. And this happened over some distance. Del Tex. If the box has some mass m and the floor has some friction force mieux que then we can assume that what happened was that friction actually took energy out of this system to the point where the kinetic energy became zero jewels didn't do it by removing mass. It did it by removing velocity. We know that work is equal to a change in kinetic energy. We also know that work is equal to F Delta X Co sign data. Okay, so let's take a look here, then, at what the math might be Now our force in this case are Onley Force is going to be FM UK, which we know from the definition is mu k times normal force as long as there's nothing else pushing it down. If we were to draw free body diagram, we have m g f N and Esme UK. This means that effin is going to be equal to M G. And we can go ahead and plug in Over here. We have, um UK times, MGI times, Delta X now cosine of the angle between them. Notice that Delta X is to the right while friction is to the left. That means cosine of theta will be co sign of 180 degrees, which is a negative one masked be equal to one half m. The final squared minus one half m the initial squared. This will be zero, and we're left with negative mu K m G Delta X is equal to one half m being not squared. So we could figure out, for example, how far this object will slide. We're going to find that the EMS cancel and that Delta X dropped a negative. Here. The negatives cancel the EMS cancel and we'll find that Delta X is equal to 1/2 mu g times be not squared. Checking the units. We have meter squared per second squared, divided by meters per second squared. So we come up with units of meters. All right, so, um, we're able to find the distance that this object would have slid. Now we can check this with some simple math of our own from Dynamic. So in dynamics, we would have said I know that F MEU is equal to mass times acceleration. I know that F MEU is also equal to Mu k M G. She's equal to M A T m's would cancel. I now know my acceleration. I know that Delta X is equal to V not t plus one half A T squared. I also know that V squared is equal to be not squared plus to a Delta X. So in order to find Delta X, I would pick this second equation and I would find that Delta X is equal to V final squared, which is zero might have minus the initial squared over to a where a is equal to mu k times G, which is exactly what I found here. Now I have a negative should recognize that in Kenya Matics I would have said that the acceleration is negative because it's to the left. So I should have had the negative in there. That's my mistake. Okay, so we come up with the exact same result using Dynamics and Kinnah Matics as we do using work and energy. Now, sometimes one will be faster than the other, and it's important to you to practice these problems so you can distinguish between when you should use one and when you should use the other. Generally speaking, when you're given a problem during this unit in your class, I recommend that you use work and energy or but if you want to check it, you can always go back and try to use forces in Kenema ticks.

Potential Energy

Equilibrium and Elasticity

Energy Conservation

Moment, Impulse, and Collisions

Rotation of Rigid Bodies

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