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(II) An exceptional standing jump would raise a person 0.80 $\mathrm{m}$off the ground. To do this, what force must a 68 -kg personexert against the ground? Assume the person crouches adistance of 0.20 $\mathrm{m}$ prior to jumping, and thus the upward forcehas this distance to act over before he leaves the ground.

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$3300 \mathrm{N}$

Physics 101 Mechanics

Chapter 4

Dynamics: Newton's Laws of Motion

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

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(II) An exceptional standi…

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(III) From what maximum he…

So let's draw the free by the diagram for the man going up would be the force sub p with 1/4 of the push. Going down would be the the man's weight. So we know the velocity that the person must have when losing contact with the ground eyes found from the equation to 12 c. We can use the acceleration due to gravity and then we're going to simply solve for the initial acceleration. Knowing that the final acceleration is zero so we can save the final squared equals the initial squared plus two g times Delta acts and here where it is gonna be zero. Therefore, V initial would be equal to the square root of negative to G Dr X, Let's solve So should the square root of negative two times negative 9.80 meters per second squared again, then multiplied by Delta X of 0.80 meters on, we find that the initial is gonna be equal to 3.960 meters per second. Not this point. We know that this velocity is the velocity that the jumper must have as a result of pushing with their legs. So we're going to use that in order to find Ah, the acceleration that the jumper missed had during their push on the floor. So we can say that again. VI final squared equals b initial squared. However, here Ah, the final squared is not gonna be equal to zero. And we can say that a The acceleration would then be equal to the final squared, uh, divided by two times Delta X knowing that the initial velocity is zero. So this would be equal to 3.960 meters per second quantity squared, divided by two times the duration. That is rather the it's essentially the the how far he moves in the ex direction when he's jumping. So this would be equal to 0.20 meters and this would equal 39.20 meters per second squared. So take note of this week and then find the accelerator. We can use the acceleration to find the magnitude of the pushing force against the ground. So we can say that the sigma okay, after some of forces in the UAE, directions would equal the force of the push minus mg and this would equal m a Therefore, the force of the push is simply gonna be able to the mass of the person multiplied by the acceleration due to gravity plus a And so this was equal 68 kilograms multiplied by 9.80 plus the acceleration that we just found 39 0.20 And then the units would, of course, being meters per second squared. And so the force of the push must be equal to approximately 3300 units. We have to round 22 significant figures because his masses only 222 significant figures. That is the end of the solution. Thank you for watching.

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