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Cornell University

University of Michigan - Ann Arbor

University of Washington

Hope College

03:02

Averell H.

(II) Superman must stop a 120-km/h train in 150 m to keep it from hitting a stalled car on the tracks. If the train's mass is $3.6 \times 10^5$ kg how much force must he exert? Compare to the weight of the train (give as %). How much force does the train exert on Superman?

04:39

Muhammed S.

(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?

03:04

Kai C.

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come to our fourth example video looking at restoring and dissipated forces in this video, we're going to consider what happens as a box slides down a ramp that has some friction on it. And again, we'll be able to analyze this in terms of dynamics with some Kinnah Matics or we can analyze it in terms of work and kinetic energy. Okay, so looking at this, then we say, All right, I know that work is equal to Delta K. Or rather, I should say total work is equal to Delta K and I will. No work done by an individual force is equal to the amount of that force Delta X, go Santa. If I were to plug in the net force, I would find the total work. All right, let's go ahead and draw a free body diagram to start because we still need to worry about forces. So we have m g. We have normal force. Let's go ahead and drawer silhouette in, and then we'll have a friction force that is anti parallel to the direction of the box. Okay, so what do we want to do here? Well, I know I have these equations and looking down here. If I were to write out my X and Y Newton's second law equations, I would find the some of forces in the Y direction. Remember, this is Thatta would be equal to FM minus m g cosine theta C equal to zero. Because it's not moving perpendicular to the inclined plane and then in the Y direction, I would have mg sine theta minus f meu K is equal to mass times acceleration. F Meu K is equal to mu k times FN, which means we can then write mg sine theta minus mu k times mg, cosine theta. It's equal to mass times acceleration. We could then take this from here and plug it into cinematics and find what RV Final is. For example, on the other hand, up here, I could say, Well, let's go ahead and try and find my the final Using this I know I have a Delta K. Let's say that the initial is equal to zero meters per second. That means Delta K is simply one half m. The final squared. If I want to plug in my F net here in the Y direction, that would be m g sine theta minus mu k m g cosine theta almost applied by Delta X. In this case, net forces in the same direction is displacement means coastline data is one. And then I said all this equal to one half MV squared. And once I've done that, I consol for Viv Final very quickly. Similarly, over here I could take this acceleration and plug it into the equation. V Final squared is equal to Vienna school squared plus to a Delta X. If you finish up the algebra here, then you'll find that plugging this acceleration into this equation we'll give you the same result as this equation up here. All right, so once again, we see that the two are equivalent. On the other hand, we could say, Well, I want to know how much work did the friction do? Well, the work done by friction would be equal to F immune times, Delta X times, the cosine of the angle between them and this will not be equal to Delta K because there's another force involved here, which is gravity now plugging it in. Though I know the F meu K is equal to Mu K mg cosine theta Delta X is still Delta X, and then the angle between the two well f m u is up the ramp and Delta X is down the ramps. That's a negative one. And so I could use this to calculate the amount of work done by the friction. Notice that looking up here, if I were to multiply the Delta X through, it would really look like that. I'm seeing the work done by gravity and then I'm removing from it the work done by friction. Okay, so we have a work total is equal to work done by gravity, minus the work done by friction.

Potential Energy

Equilibrium and Elasticity

Energy Conservation

Moment, Impulse, and Collisions

Rotation of Rigid Bodies

02:15

02:42

04:42

02:20